Answer:
[tex] \boxed{3q(4 {p}^{2} - 3q)}[/tex]Step-by-step explanation:
[tex] \mathsf{ 12 {p}^{2} q - 9 {q}^{2} }[/tex]
In such an expression, the factor which is present in all terms of the expression is taken out as common and each term of the expression should be divided by the common factor to get another factor.
Factor out 3q from the expression
[tex] \mathsf{ = 3q(4 {p}^{2} - 3q)}[/tex]
Hope I helped!
Best regards!
Factorization of 12p²q-9q² is 3q(4p²-3q).
What is Factorization?Factorization is defined as breaking an entity into a product of another entity, or factors, which when multiplied together give the original number.
Here, given expression is, 12p²q-9q²
Now, by factorizing this we get,
3q(4p²-3q)
Hence, required factorization is 3q(4p²-3q)
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Which of the following best describes the graph shown below?
16
A1
1
14
O A This is the graph of a linear function
B. This is the graph of a one-to-one function
C. This is the graph of a function, but it is not one to one
D. This is not the graph of a function
The vertical line test helps us see that we have a function. Note how it is not possible to draw a single straight line through more than one point on the curve. Any x input leads to exactly one y output. This graph passes the vertical line test. Therefore it is a function.
The function is not one-to-one because the graph fails the horizontal line test. Here it is possible to draw a single straight horizontal line through more than one point on the curve. The horizontal line through y = 2 is one example of many where the graph fails the horizontal line test, meaning the function is not one-to-one.
The term "one-to-one" means that each y value only pairs up with one x value. Here we have something like y = 2 pair up with multiple x values at the same time. This concept is useful when it comes to determining inverse functions.
Which equation does the graph of the systems of equations solve? 2 linear graphs. They intersect at 1,4
Answer:
See below.
Step-by-step explanation:
There is an infinite n umber of systems of equations that has (1, 4) as its solution. Are you given choices? Try x = 1 and y = 4 in each equation of the choices. The set of two equations that are true when those values of x and y are used is the answer.
Complete the square to transform the expression x2 - 2x - 2 into the form a(x - h)2 + k
Answer:
A
Step-by-step explanation:
Find the vertex form of the quadratic function below.
y = x^2 - 4x + 3
This quadratic equation is in the form y = a{x^2} + bx + cy=ax
2
+bx+c. However, I need to rewrite it using some algebraic steps in order to make it look like this…
y = a(x - h)^2 + k
This is the vertex form of the quadratic function where \left( {h,k} \right)(h,k) is the vertex or the “center” of the quadratic function or the parabola.
Before I start, I realize that a = 1a=1. Therefore, I can immediately apply the “completing the square” steps.
STEP 1: Identify the coefficient of the linear term of the quadratic function. That is the number attached to the xx-term.
STEP 2: I will take that number, divide it by 22 and square it (or raise to the power 22).
STEP 3: The output in step #2 will be added and subtracted on the same side of the equation to keep it balanced.
Think About It: If I add 44 on the right side of the equation, then I am technically changing the original meaning of the equation. So to keep it unchanged, I must subtract the same value that I added on the same side of the equation.
STEP 4: Now, express the trinomial inside the parenthesis as a square of a binomial, and simplify the outside constants.
After simplifying, it is now in the vertex form y = a{\left( {x - h} \right)^2} + ky=a(x−h)
2
+k where the vertex \left( {h,k} \right)(h,k) is \left( {2, - 1} \right)(2,−1).
Visually, the graph of this quadratic function is a parabola with a minimum at the point \left( {2, - 1} \right)(2,−1). Since the value of “aa” is positive, a = 1a=1, then the parabola opens in upward direction.
Example 2: Find the vertex form of the quadratic function below.
The approach to this problem is slightly different because the value of “aa” does not equal to 11, a \ne 1a
=1. The first step is to factor out the coefficient 22 between the terms with xx-variables only.
STEP 1: Factor out 22 only to the terms with variable xx.
STEP 2: Identify the coefficient of the xx-term or linear term.
STEP 3: Take that number, divide it by 22, and square.
STEP 4: Now, I will take the output {9 \over 4}
4
9
and add it inside the parenthesis.
By adding {9 \over 4}
4
9
inside the parenthesis, I am actually adding 2\left( {{9 \over 4}} \right) = {9 \over 2}2(
4
9
)=
2
9
to the entire equation.
Why multiply by 22 to get the “true” value added to the entire equation? Remember, I factored out 22 in the beginning. So for us to find the real value added to the entire equation, we need to multiply the number added inside the parenthesis by the number that was factored out.
STEP 5: Since I added {9 \over 2}
2
9
to the equation, then I should subtract the entire equation by {9 \over 2}
2
9
also to compensate for it.
STEP 6: Finally, express the trinomial inside the parenthesis as the square of binomial and then simplify the outside constants. Be careful combining the fractions.
It is now in the vertex form y = a{\left( {x - h} \right)^2} + ky=a(x−h)
2
+k where the vertex \left( {h,k} \right)(h,k) is \left( {{{ - \,3} \over 2},{{ - 11} \over 2}} \right)(
2
−3
,
2
−11
).
Example 3: Find the vertex form of the quadratic function below.
Solution:
Factor out - \,3−3 among the xx-terms.
The coefficient of the linear term inside the parenthesis is - \,1−1. Divide it by 22 and square it. Add that value inside the parenthesis. Now, figure out how to make the original equation the same. Since we added {1 \over 4}
4
1
inside the parenthesis and we factored out - \,3−3 in the beginning, that means - \,3\left( {{1 \over 4}} \right) = {{ - \,3} \over 4}−3(
4
1
)=
4
−3
is the value that we subtracted from the entire equation. To compensate, we must add {3 \over 4}
4
3
outside the parenthesis.
Therefore, the vertex \left( {h,k} \right)(h,k) is \left( {{1 \over 2},{{11} \over 4}} \right)(
2
1
,
4
11
).
Example 4: Find the vertex form of the quadratic function below.
y = 5x^2 + 15x - 5
Solution:
Factor out 55 among the xx-terms. Identify the coefficient of the linear term inside the parenthesis which is 33. Divide it by 22 and square to get {9 \over 4}
4
9
.
Add {9 \over 4}
4
9
inside the parenthesis. Since we factored out 55 in the first step, that means 5\left( {{9 \over 4}} \right) = {{45} \over 4}5(
4
9
)=
4
45
is the number that we need to subtract to keep the equation unchanged.
Express the trinomial as a square of binomial, and combine the constants to get the final answer.
Therefore, the vertex \left( {h,k} \right)(h,k) is {{ - \,3} \over 2},{{ - \,65} \over 4}
2
−3
,
4
−65
.
Answer:
(x - 1 )^2 - 3
Step-by-step explanation:
( x - 1 )^2 + ( -3)
x^2 - 2x + 1 - 3
x^2 - 2x - 2
1. Suzette ran and biked for a total of 80 miles in 9 hours. Her average running speed was 5 miles per hour (mph) and her average biking speed was 12 mph. Let x = total hours Suzette ran. Let y = total hours Suzette biked. Use substitution to solve for x and y. Show your work. Check your solution. (a) How many hours did Suzette run? (b) How many hours did she bike?
Answer:
a) Suzette ran for 4 hours
b) Suzette biked for 5 hours
Step-by-step explanation:
Speed is rate of distance traveled, it is the ratio of distance traveled to time taken. It is given by:
Speed = distance / time
The total distance ran and biked by Suzette (d) = 80 miles, while the total time ran and biked by Suzette (t) = 9 hours.
For running:
Her speed was 5 miles per hour, let the total hours Suzette ran be x and the total distance she ran be p, hence since Speed = distance / time, therefore:
5 = p / x
p = 5x
For biking:
Her speed was 12 miles per hour, let the total hours Suzette ran be y and the total distance she ran be q, hence since Speed = distance / time, therefore:
12 = q / y
q = 12y
The total distance ran and biked by Suzette (d) = Distance biked + distance ran
d = p + q
80 = p + q
80 = 5x + 12y (1)
The total time taken to run and bike by Suzette (t) = time spent to bike + time spent to run
t = x + y
9 = x + y (2)
Solving equation 1 and equation 2, multiply equation 2 by 5 and subtract from equation 1:
7y = 35
y = 35/7
y = 5 hours
Put y = 5 in equation 2:
9 = x + 5
x = 9 -5
x = 4 hours
a) Suzette ran for 4 hours
b) Suzette biked for 5 hours
What is 20 to 7 minus 1 hour 40 mins Will award brainliest
6:40 or 6 hour 40 minutes,
if you go back(subtract) 1 hour and 40 minutes
i.e. 6hours 40 minutes- 1 hour 40 minutes
subtract minutes from minutes and hours from hours,
5:00
note that here the minutes value is not negative so it was not a problem, what If it was 6:40-1:50?
Two co-interior angles
formed between the
two parallel lines are in the ratio of 11.7.
Find the measures
of angles
Answer:
110° and 70°
Step-by-step explanation:
The angles are supplementary, thus sum to 180°
sum the parts of the ratio, 11 + 7 = 18
divide 180° by 18 to find the value of one part of the ratio
180° ÷ 18 = 10° ← value of 1 part of the ratio
Thus
11 parts = 11× 10° = 110°
7 parts = 7 × 10° = 70°
The angles are 110° and 70°
I need help asap!!!
AB =
Round your answer to the nearest hundredth.
B
?
2
25°
С
A
Answer:
? = 4.73
Step-by-step explanation:
Since this is a right triangle we can use trig functions
sin theta = opp / hyp
sin 25 = 2 / ?
? sin 25 = 2
? = 2 / sin 25
? =4.732403166
To the nearest hundredth
? = 4.73
3/4a−16=2/3a+14 PLEASE I NEED THIS QUICK and if you explain the steps that would be geat:) Thank youuuuuuu
Answer:
360
Step-by-step explanation:
3/4a - 16 = 2/3a + 14 ⇒ collect like terms 3/4a - 2/3a = 14 + 16 ⇒ bring the fractions to same denominator9/12a - 8/12a = 30 ⇒ simplify fraction1/12a = 30 ⇒ multiply both sides by 12a = 30*12a = 360 ⇒ answerI need help on this :(
Answer:
26⁹
Step-by-step explanation:
26 * 26⁸
= 26¹ * 26⁸
= 26¹⁺⁸
= 26⁹
Consider the following system of equations: y=2x−2 6x+3y=2 The graph of these equations consists of two lines that: 1. intersect at more than one point. 2. intersect in an infinite number of points. 3. intersect at exactly one point. 4. do not intersect.
Answer:
3. Intersect at exactly one point. ( (2/3), (-2/3) )
Step-by-step explanation:
To make the comparison of these lines easier, let's rewrite the 2nd equation into slope-intercept form, as the 1st equation is in slope-intercept form.
[1] y = 2x - 2
---------------------
[2] 6x + 3y = 2 ==> 3y = 2 - 6x ==> y = -2x + (2/3)
[2] y = -2x + (2/3)
So now that we have both equations in slope-intercept form, we can see that the two equations are both linear, have different slopes, and have different y-intercepts.
Since these equations have both different slopes and different y-intercepts, we know that the lines will cross at least one point. We can confirm that the lines only cross at a single point using the fact that both equations are linear, meaning there will only be one point of crossing. To find that point, we can simply set the equations equal to each other.
y = 2x - 2
y = -2x + (2/3)
2x - 2 = -2x + (2/3)
4x = (8/3)
x = (8/12) = (2/3)
And plug this x value back into one of the equations:
y = 2x - 2
y = 2(2/3) - 2
y = (4/3) - (6/3)
y = (-2/3)
Thus these lines only cross at the point ( (2/3), (-2/3) ).
Cheers.
Answer:
I don't understand the question
prove tan(theta/2)=sin theta/1+cos theta for theta in quadrant 1 by filling in the calculations and reasons. PLEASE HELP!!!!
Answer:
See explanation
Step-by-step explanation:
We have to prove the identity
[tex]tan(\frac{\Theta }{2})=\frac{sin\Theta}{1+cos\Theta }[/tex]
We will take right hand side of the identity
[tex]\frac{sin\Theta}{1+cos\Theta}=\frac{2sin(\frac{\Theta }{2})cos(\frac{\Theta }{2})}{1+[2cos^{2}(\frac{\Theta }{2})-1]}[/tex]
[tex]=\frac{2sin(\frac{\Theta }{2})cos(\frac{\Theta }{2})}{2cos^{2}(\frac{\Theta }{2})}=\frac{sin(\frac{\Theta }{2})}{cos(\frac{\Theta }{2})}[/tex]
[tex]=tan(\frac{\Theta }{2})[/tex] [ Tan θ will be positive since θ lies in 1st quadrant ]
how do you solve 2m-10=44+8m
Answer:
m = -9
Step-by-step explanation:
2m-10=44+8m
Subtract 2m from each side
2m-2m-10=44+8m-2m
-10 = 44+6m
Subtract 44 from each side
-10-44 = 44-44+6m
-54 = 6m
Divide by 6
-54/6 = 6m/6
-9 = m
Answer:
solve by solving the salvation for equation don't be a slave get educated from what's gave
The graph below shows Roy's distance from his office (y), in miles, after a certain amount of time (x), in minutes: Graph titled Roys Distance Vs Time shows 0 to 10 on x and y axes at increments of 1.The label on x axis is time in minutes and that on y axis is Distance from Office in miles. Lines are joined at the ordered pairs 0, 0 and 1, 1 and 2, 2 and 3, 3 and 4, 4 and 5, 4 and 6, 4 and 7, 4.5 and 7.5, 5 and 8, 6. Four students described Roy's motion, as shown in the table below: Student Description Peter He drives a car at a constant speed for 4 minutes, then stops at a crossing for 6 minutes, and finally drives at a variable speed for the next 2 minutes. Shane He drives a car at a constant speed for 4 minutes, then stops at a crossing for 2 minutes, and finally drives at a variable speed for the next 8 minutes. Jamie He drives a car at a constant speed for 4 minutes, then stops at a crossing for 6 minutes, and finally drives at a variable speed for the next 8 minutes. Felix He drives a car at a constant speed for 4 minutes, then stops at a crossing for 2 minutes, and finally drives at a variable speed for the next 2 minutes. Which student most accurately described Roy's motion? Peter Shane Jamie Felix
Answer:
Felix
Step-by-step explanation:
The graph contains 3 segments,
first one is for the first 4minutes,
second one is for the next 2 minutes (standing still)
third one is for the last 2 minutes.
Only Felix has it right, the other students use absolute time in their statements, in stead of the difference between start and end. (e.g., from 4 to 6 is 2 minutes).
The student that most accurately described Roy's motion is Felix.
How to find the function which was used to make graph?There are many tools we can use to find the information of the relation which was used to form the graph.
A graph contains data of which input maps to which output.
Analysis of this leads to the relations which were used to make it.
We need to find the student that most accurately described Roy's motion.
Here we can see that the graph contains 3 segments, first one is for the first 4 minutes, Second one is for the next 2 minutes (standing still) and the third one is for the last 2 minutes.
Now, Only Felix has it right, the other students use absolute time in their statements, in stead of the difference between start and end.
Therefore, the student that most accurately described Roy's motion is Felix.
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a diagonal of rectangle forms a 30 degree angle with each of the longer sides of the rectangle. if the length of the shorter side is 3, what is the length of the diagonal
Answer:
Length of diagonal = 6
Step-by-step explanation:
Given that
Diagonal of a rectangle makes an angle of [tex]30^\circ[/tex] with the longer side.
Kindly refer to the attached diagram of the rectangle ABCD such that diagonal BD makes angles of [tex]30^\circ[/tex] with the longer side CD and BA.
[tex]\angle CDB =\angle DBA =30^\circ[/tex]
Side AD = BC = 3 units
To find:
Length of diagonal BD = ?
Solution:
We can use the trigonometric ratio to find the diagonal in the [tex]\triangle BCD[/tex] because [tex]\angle C =90^\circ[/tex]
Using the sine :
[tex]sin\theta = \dfrac{Perpendicular }{Hypotenuse }[/tex]
[tex]sin\angle CDB = \dfrac{BC}{BD}\\\Rightarrow sin30^\circ = \dfrac{3}{BD}\\\Rightarrow \dfrac{1}2 = \dfrac{3}{BD}\\\Rightarrow BD =2 \times 3 \\\Rightarrow BD = \bold{6 }[/tex]
So, the answer is:
Length of diagonal = 6
Shaquira is baking cookies to put in packages for a fundraiser. Shaquira has made 86 8686 chocolate chip cookies and 42 4242 sugar cookies. Shaquira wants to create identical packages of cookies to sell, and she must use all of the cookies. What is the greatest number of identical packages that Shaquira can make?
Answer: 2
Step-by-step explanation:
Given: Shaquira has made 86 chocolate chip cookies and 42 sugar cookies.
Shaquira wants to create identical packages of cookies to sell, and she must use all of the cookies.
Now, the greatest number of identical packages that Shaquira can make= GCD of 86 and 42
Prime factorization of 86 and 42:
86 = 2 ×43
42 = 2 × 3 × 7
GCD of 86 and 42 = 2 [GCD = greatest common factor]
Hence, the greatest number of identical packages that Shaquira can make =2
I need hellp please its my last chance to become a senior please someone
Answer:
d= 6
r= 6/2
r=3
V= π. r². h
V= π . 3². 14
V= π. 9 . 14
V= π 126 cm³
V= 126 π cm³ (π not in number)
hope it helps^°^
Answer:if you use the formula it is 126 pi cm cubed
The answer is c
Step-by-step explanation:
An entomologist is studying the reproduction of ants. If an ant colony started with 50 ants, and each day, their population increases by 10%, how many ants will be in the colony 5 days later? *
Step-by-step explanation: Ants are one of the most abundant insects on our planet and the reasons are their eusocial, complex societal behaviors and their ability to survive in many and various ecosystems. Like most other animal societies, reproduction is one of the core reasons why ants are so prevalent.
Acrobat Ant
Reproduction for ants is a complex phenomenon that involves finding, selecting and successfully fertilizing females to ensure that the eggs laid are able to survive and molt through the successive stages of the ant’s life cycle – larvae, pupae and adults.
Answer:
81
Step-by-step explanation:
Start: 50
After 1 day: 50 * 1.1
After 2 days: 50 * 1.1 * 1.1 = 50 * 1.1^2
After 3 days: 50 * 1.1^2 * 1.1 = 50 * 1.1^3
...
After 5 days: 50 * 1.1^5 = 80.53
Answer: 81
PLEASE help me with this question! No nonsense answers please. This is really urgent.
Answer:
last option
Step-by-step explanation:
Let's call the original angle x° and the radius of the circle y. The area of the original sector would be x / 360 * πy². The new angle, which is a 40% increase from x, can be represented as 1.4x so the area of the new sector is 1.4x / 360 * πy². Now, to find the corresponding change, we can calculate 1.4x / 360 * πy² ÷ x / 360 * πy² = (1.4x / 360 * πy²) * (360 * πy² / x). 360 * πy² cancels out so we're left with 1.4x / x which becomes 1.4, signifying that the area of the sector increases by 40%.
PLEaSE HELP!!!!!! will give brainliest to first answer
Answer:
The coordinates of A'C'S'T' are;
A'(-7, 2)
C'(-9, -1)
S'(-7, -4)
T'(-5, -1)
The correct option is;
B
Step-by-step explanation:
The coordinates of the given quadrilateral are;
A(-3, 1)
C(-5, -2)
S(-3, -5)
T(-1, -2)
The required transformation is T₍₋₄, ₁₎ which is equivalent to a movement of 4 units in the leftward direction and 1 unit upward
Therefore, we have;
A(-3, 1) + T₍₋₄, ₁₎ = A'(-7, 2)
C(-5, -2) + T₍₋₄, ₁₎ = C'(-9, -1)
S(-3, -5) + T₍₋₄, ₁₎ = S'(-7, -4)
T(-1, -2) + T₍₋₄, ₁₎ = T'(-5, -1)
Therefore, the correct option is B
Will give Brainliest, Please show work.
Answer:
Hi, there!!
Hope you mean the answers in the solution.
Hope it helps...
Answer:
Step-by-step explanation:
7)
JKLM is a isosceles trapezium.
KL // JM
∠K + ∠J = 180 {Co interior angles}
50 +∠J = 180
∠J = 180 - 50
∠J = 130
As it is isosceles, non parallel sides KJ = LM &
∠L = ∠K
∠L = 50
∠M = ∠J
∠M = 130
8)JKLM is a isosceles trapezium.
KL // JM
∠K + ∠J = 180 {Co interior angles}
100 +∠J = 180
∠J = 180 - 100
∠J = 80
As it is isosceles, non parallel sides KJ = LM &
∠L = ∠K
∠L = 100
∠M = ∠J
∠M = 80
Given the equations of a straight line f(x) (in slope-intercept form) and a parabola g(x) (in standard form), describe how to determine the number of intersection points, without finding the coordinates of such points. Do not give an example.
Answer:
Step-by-step explanation:
Hello, when you try to find the intersection point(s) you need to solve a system like this one
[tex]\begin{cases} y&= m * x + p }\\ y &= a*x^2 +b*x+c }\end{cases}[/tex]
So, you come up with a polynomial equation like.
[tex]ax^2+bx+c=mx+p\\\\ax^2+(b-m)x+c-p=0[/tex]
And then, we can estimate the discriminant.
[tex]\Delta=(b-m)^2-4*a*(c-p)[/tex]
If [tex]\Delta<0[/tex] there is no real solution, no intersection point.
If [tex]\Delta=0[/tex] there is one intersection point.
If [tex]\Delta>0[/tex] there are two real solutions, so two intersection points.
Hope this helps.
I need help fast please
Answer:
Difference : 4th option
Step-by-step explanation:
The first thing we want to do here is to factor the expression x² + 3x + 2. This will help us if it is similar to the factored expression " ( x + 2 )( x + 1 ). " The denominators will be the same, and hence we can combine the fractions.
x² + 3x + 2 - Break the expression into groups,
( x² + x ) + ( 2x + 2 ) - Factor x from x² + x and 2 from 2x + 2,
x( x + 1 ) + 2( x + 2 ) - Group,
( x + 2 )( x + 1 )
This is the same as the denominator of the other fraction, and therefore we can combine the fractions.
x - 1 / ( x + 2 )( x + 1 )
As you can see this is not any of the options present, as we have not expanded ( x + 2 )( x + 1 ). Remember previously that ( x + 2 )( x + 1 ) = x² + 3x + 2. Hence our solution is x - 1 / x² + 3x + 2, or option d.
The equation of line WX is 2x + y = −5. What is the equation of a line perpendicular to line WX in slope-intercept form that contains point (−1, −2)?
Answer: [tex]y=\dfrac12x-\dfrac{3}{4}[/tex]
Step-by-step explanation:
Given, The equation of line WX is 2x + y = −5.
It can be written as [tex]y=-2x-5[/tex] comparing it with slope-intercept form y=mx+c, where m is slope and c is y-intercept, we have
slope of WX = -2
Product of slopes of two perpendicular lines is -1.
So, (slope of WX) × (slope of perpendicular to WX)=-1
[tex]-2\times\text{slope of WX}=-1\\\\\Rightarrow\ \text{slope of WX}=\dfrac{1}{2}[/tex]
Equation of a line passes through (a,b) and has slope m:
[tex]y-b=m(x-a)[/tex]
Equation of a line perpendicular to WX contains point (−1, −2) and has slope [tex]=\dfrac12[/tex]
[tex]y-(-2)=\dfrac{1}{2}(x-(-1))\\\\\Rightarrow\ y+2=\dfrac12(x+1)\\\\\Rightarrow\ y+2=\dfrac12x+\dfrac12\\\\\Rightarrow\ y=\dfrac12x+\dfrac12-2\\\\\Rightarrow\ y=\dfrac12x-\dfrac{3}{4}[/tex]
Equation of a line perpendicular to line WX in slope-intercept form that contains point (−1, −2) [tex]:y=\dfrac12x-\dfrac{3}{4}[/tex]
SAVINGS ACCOUNT Demetrius deposits $120 into his account. One week later, he withdraws $36. Write an addition expression to represent this situation. How much higher or lower is the amount in his account after these two transactions?
Answer:
+$120 - $36
Higher by $84
Step-by-step explanation:
Addition expression is an equation without the equals to sign
$120 - $36
When the first expression was made, the account was higher by $120
After the second transaction, the account would be higher by $120 - $36 = $84
PLEASE HELP
Find the area and the perimeter of the shaded regions below. Give your answer as a completely simplified exact value in terms of π (no approximations). The figures below are based on semicircles or quarter circles and problems b), c), and d) are involving portions of a square.
Answer:
perimeter is 4 sqrt(29) + 4pi cm
area is 40 + 8pi cm^2
Step-by-step explanation:
We have a semicircle and a triangle
First the semicircle with diameter 8
A = 1/2 pi r^2 for a semicircle
r = d/2 = 8/2 =4
A = 1/2 pi ( 4)^2
=1/2 pi *16
= 8pi
Now the triangle with base 8 and height 10
A = 1/2 bh
=1/2 8*10
= 40
Add the areas together
A = 40 + 8pi cm^2
Now the perimeter
We have 1/2 of the circumference
1/2 C =1/2 pi *d
= 1/2 pi 8
= 4pi
Now we need to find the length of the hypotenuse of the right triangles
using the pythagorean theorem
a^2+b^2 = c^2
The base is 4 ( 1/2 of the diameter) and the height is 10
4^2 + 10 ^2 = c^2
16 + 100 = c^2
116 = c^2
sqrt(116) = c
2 sqrt(29) = c
Each hypotenuse is the same so we have
hypotenuse + hypotenuse + 1/2 circumference
2 sqrt(29) + 2 sqrt(29) + 4 pi
4 sqrt(29) + 4pi cm
Step-by-step explanation:
First we need to deal with the half circle. The radius of this circle is 4, because the diameter is 8. The formula for the circumference of a circle is 2piR.
2pi4 so the perimeter for the half circle would be 8pi/2.
The area of that half circle would be piR^2 so 16pi/2.
Now moving on the triangle part, we need to find the hypotenuse side of AC. We will use the pythagoram theorem. 4^2+10^2=C^2
16+100=C^2
116=C^2
C=sqrt(116)
making the perimeter of this triangle 2×sqrt(116)
The area of this triangle is 8×10=80, than divided by 2 which is equal to 40.
We than just need to add up the perimeters and areas for both the half circle and triangle.
The area would be equal to 8pi+40
The perimeter would be equal to 4pi+4(sqrt(29))
How do u simplify each expression by combining like terms?
Answer:
1. 8y - 9y = -1y
( 8 - 9 = -1)
3. 8a - 6 +a - 1
( i have showed the like terms here)
8a - 1a= 7a
-6 - 1 = -7
7a - 7
5. -x - 2 + 15x
( i have showed the like terms here)
-x + 15x = 14x
(x = 1)
14x + 2
7. 8d - 4 - d - 2
( i have showed the like terms here)
8d - d = 7d
-4 -2 = -6
7d - 6
8. 9a + 8 - 2a - 3 - 5a
( i have showed the like terms here)
9a - 2a - 5a = 2a
8 - 3= 5
2a + 5
Find the vertex of f(x)= x^2+ 6x + 36
Pls help soon
Answer:
vertex(-3,27)
Step-by-step explanation:
f(x)= x^2+ 6x + 36 ( a=1,b=6,c=36)
V(h,k)
h=-b/2a=-6/2=-3
k=f(-3)=3²+6(-3)+36
f(-3)=9-18+36=27
vertex(-3,27)
What is [tex]3^2*3^5[/tex]?
Answer:
[tex]3^7[/tex]
Step-by-step explanation:
[tex]3^2*3^5[/tex]
[tex]\text {Apply Product Rule: } a^b+a^c=a^{b+c}\\\\3^2*3^5=3^{2+5}=3^7[/tex]
State whether the given measurements determine zero, one, or two triangles. A = 58°, a = 25, b = 28
Answer:
1
Step-by-step explanation:
I believe it is 1. Just picture or draw a diagram of the constraints. Don't quote me on this though...
Answer:
Step-by-step explanation:
apply sine formula
[tex]\frac{a}{sin ~A} =\frac{b}{sin~B} \\\frac{25}{sin~58} =\frac{28}{sin ~B} \\sin~B=\frac{28}{25} \times sin~58\\B=sin^{-1} (\frac{28}{25} \times sin ~58)=71.77 \approx 72 ^\circ[/tex]
so third angle=180-(58+72)=180-130=50°
∠C=50°
[tex]cos ~C=\frac{a^2+b^2-c^2}{2ab} \\or ~2abcos~C=a^2+b^2-c^2\\2*25*28*cos ~50=25^2+28^2-c^2\\c^2=625+784-1400 *cos~50\\c^2=1409-899.90\\c^2=509.1\\c=\sqrt{509.1} \approx 22.56 \approx 22.6[/tex]
so one triangle is formed.
