Answers
Step-by-step explanation:
You look for the common factor of both of them which in this case is 5, therefore it's
5(x+7)..just divide 5 in 5x and in 35
Related Questions
12) Find the angles between 0o and 360o where sec θ = −3.8637 . Round to the nearest 10th of a degree:
Please show all work
Answers
9514 1404 393
Answer:
105.0°, 255.0°
Step-by-step explanation:
Many calculators do not have a secant function, so the cosine relation must be used.
sec(θ) = -3.8637
1/cos(θ) = -3.8637
cos(θ) = -1/3.8637
θ = arccos(-1/3.8637) ≈ 105.000013°
The secant and cosine functions are symmetrical about the line θ = 180°, so the other solution in the desired range is ...
θ = 360° -105.0° = 255.0°
The angles of interest are θ = 105.0° and θ = 255.0°.
Which expression defines the given series for seven terms?
–4 + (–5) + (–6) + . . .
Answers
Answer: -n+(-n-1)
Step-by-step explanation:
Expression will be -n + (-1)
Series
-4 +(-5)+(-6)+(-7)+(-8)+(-9)+(-10)+(-11)+(-12)+(-13) and so on
Here number -n has + (-n-1) being added to it
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Which statement is true about the ratios of squares to
cicles in the tables? PLS HURRY!!!!
Answers
Answer:
show us a screenshot or image
or type it out, copy paste
Step-by-step explanation:
Translate this phrase into an algebraic expression.
61 less than twice Jenny's age
Use the variable j to represent Jenny's age.
Answers
j = Jenny's age
use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x
Answers
First check the characteristic solution: the characteristic equation for this DE is
r ² - 3r + 2 = (r - 2) (r - 1) = 0
with roots r = 2 and r = 1, so the characteristic solution is
y (char.) = C₁ exp(2x) + C₂ exp(x)
For the ansatz particular solution, we might first try
y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)
where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).
However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :
y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)
Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.
y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)
… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)
y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)
… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)
Substituting every relevant expression and simplifying reduces the equation to
(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)
… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]
… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]
= 2x + (1 + 2x) exp(x) + 4 exp(3x)
… … …
2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)
= 2x + (1 + 2x) exp(x) + 4 exp(3x)
Then, equating coefficients of corresponding terms on both sides, we have the system of equations,
x : 2a = 2
1 : -3a + 2b = 0
exp(x) : 2c - d = 1
x exp(x) : -2c = 2
exp(3x) : 2e = 4
Solving the system gives
a = 1, b = 3/2, c = -1, d = -3, e = 2
Then the general solution to the DE is
y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)
Again need help with these ones I don’t understand and they have to show work
Answers
[tex]x^2 - 18x + 81= 65 + 81[/tex]
[tex](x - 9)^2 = 146[/tex]
Taking the square root of both sides, we get
[tex]x - 9 = \pm\sqrt{146}[/tex]
or
[tex]x = 9 \pm \sqrt{146} = 9 \pm 12.1 = 21.1\:\text{and}\:-3.1[/tex]
10. (30-i)-(18+6i)+30i
Answers
Answer:
[tex]12+23i[/tex]
Step-by-step explanation:
[tex](30−i)−(18+6i)+30i[/tex]
[tex]30−i−18−6i+30i[/tex]
[tex]12−i−6i+30i[/tex]
[tex]12−7i+30i[/tex]
[tex]12+23i[/tex]
Hope it is helpful....Please answer this question
Answers
A.) V’ (-3,-5), K’ (-1,-2), B’ (3,-1), Z’(2,-5)
B.) V’(-4, 1), K’(-2, 4), B(2,5) Z’ (1, 1)
C.) V’ (-3,-4), K’(-1,-1) B’ (3,0), Z’(2,-4)
D.) V’ (-1,0), K’ (1, 3), B’(5,4), Z’(4,0)
Answers
Answer:
C
Step-by-step explanation:
this is a "translation" - a shift of the object without changing its shadow and size.
this shift is described by a "vector" - in 2D space by the x and y distances to move.
we have here (1, 0) - so, we move every point one unit to the right (positive x direction) and 0 units up/down.
therefore, C is the right answer (the x coordinates of the points are increased by 1, the y coordinate are unchanged).
WORTH 100 POINTS!
The function h(x) is quadratic and h(3) = h(-10) = 0. Which could represent h(x)?
1) h(x) = x2 - 13x - 30
2) h(x) = x2 - 7x - 30
3) h(x) = 2x2 + 26x - 60
4) h(x) = 2x2 + 14x - 60
Answers
Answer:
h(x) = 2x^2 +14x -60
Step-by-step explanation:
A quadratic is of the form
h(x) = ax^2 + bx +c
h(3) = h(-10) = 0
This tells us that the zeros are at x=3 and x = -10
We can write the equation in the form
h(x) = a( x-z1)(x-z2) where z1 and z2 are the zeros
h(x) = a(x-3) (x- -10)
h(x) = a(x-3) (x+10)
FOIL
h(x) = a( x^2 -3x+10x-30)
h(x) = a(x^2 +7x -30)
Let a = 2
h(x) = 2x^2 +14x -60
It means
zeros are 3 and -10
Form equation
y=x²-(3-10)x+(-10)(3)y=x²+7x-30Multi ply by 2
y=2x²+14x-60Option D
Find the sum of ∑3/k=0 k^2
Answers
Answer:
[tex]14[/tex]
Step-by-step explanation:
Given
[tex]\displaystyle \sum_{k=0}^3k^2[/tex]
Let's break down each part. The input at the bottom, in this case [tex]k=0[/tex], is assigning an index [tex]k[/tex] at a value of [tex]0[/tex]. This is the value we should start with when substituting into our equation.
The number at the top, in this case 3, indicates the index we should stop at, inclusive (meaning we finish substituting that index and then stop). The equation on the right, in this case [tex]k^2[/tex], is the equation we will substitute each value in. After we substitute our starting index, we'll continue substituting indexes until we reach the last index, then add up each of the outputs produced.
Since [tex]k=0[/tex] is our starting index, start by substituting this into [tex]k^2[/tex]:
[tex]0^2=0[/tex]
Now continue with [tex]k=1[/tex]:
[tex]1^1=1[/tex]
Repeat until we get to the ending index, [tex]k=3[/tex]. Remember to still use [tex]k=3[/tex] before stopping!
Substituting [tex]k=2[/tex]:
[tex]2^2=4[/tex]
Substituting [tex]k=3[/tex]:
[tex]3^2=9[/tex]
Since 3 is the index we end at, we stop here. Now we will add up each of the outputs:
[tex]0+1+4+9=\boxed{14}[/tex]
Therefore, our answer is:
[tex]\displaystyle \sum_{k=0}^3k^2=0+1+4+9=\boxed{14}[/tex]
Answer:
14
Step-by-step explanation:
∑3/k=0 k^2
Let k=0
0^2 =0
Let k = 1
1^2 =1
Let k =2
2^2 = 4
Let k = 3
3^2 = 9
0+1+4+9 = 14
You flip a coin that is not fair, the prbability of heads on each flip is 0.7. if the coin shows heads, you draw a marble from urn h with 1 blue and 4 red marbles. if the coin shows tails, you draw a marble from urn t with 3 blue and 1 red marble. Find the following probabilities:
a. The probability of choosing a red marble.
b. The probability of choosing a blue marble, given that the coin showed heads.
c. The probability that the coin showed tails, given that the marble was red.
Answers
Solution :
P(H) = 0.7 ; P(T) = 0.3
If heads, then Urn H, 1 blue and 4 red marbles.
If tails, then Urn T , 3 blue and 1 red marbles.
a).
P ( choosing a Red marble )
= P (H) x P( Red from Urn H) + P (T) x P( Red from Urn T)
[tex]$=0.7 \times \frac{4}{5} + 0.3 \times \frac{1}{4}$[/tex]
= 0.56 + 0.075
= 0.635
b). If P (B, if coin showed heads)
If heads, then marble is picked from Urn H.
Therefore,
P (Blue) [tex]$=\frac{1}{5}$[/tex]
= 0.2
c). P (Tails, if marble was red)
[tex]$=P (T/R) = \frac{P(R/T)}{P(R)} \ P(T)$[/tex]
Where P (R/T) = P ( red, if coin showed tails)
[tex]$=\frac{1}{4}$[/tex]
= 0.25 (As Urn T is chosen)
P (R) = P (Red) = 0.635 (from part (a) )
P (T) = P (Tails) = 0.3
∴ [tex]$P(T/R) = \frac{0.25 \times 0.3}{0.635}$[/tex]
= 0.118
why infinity ( ) can’t be included in an inequality?
Answers
Answer:
Step-by-step explanation:
Because then the value on the other side will be unbounded by the infinity sign while expressing the answers on a number line.
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what decimal is equivalent to 0.85
Answers
Answer: 17/20
Step-by-step explanation:
0.85 = 85/100 = 17/20
The number 0.85 can be written using the fraction 85/100 which is equal to 17/20 when reduced to lowest terms.
A baseball team plays in a stadium that holds 58000 spectators. With the ticket price at $12 the average attendance has been 25000. When the price dropped to $9, the average attendance rose to 29000. Assume that attendance is linearly related to ticket price.
Required:
a. Find the demand function p(x), where x is the number of the spectators.
b. How should ticket prices be set to maximize revenue?
Answers
Answer:
We need to assume that the relationship is linear.
a) Remember that a linear relation is written as:
y = a*x + b
then we will have:
p(x) = a*x + b
where a is the slope and b is the y-intercept.
If we know that the line passes through the points (a, b) and (c, d), then the slope can be written as:
y = (d - b)/(c - a)
In this case, we know that:
if the ticket has a price of $12, the average attendance is 25,000
Then we can define this with the point:
(25,000 , $12)
We also know that when the price is $9, the attendance is 29,000
This can be represented with the point:
(29,000, $9)
Then we can find the slope as:
a = ($9 - $12)/(29,000 - 25,000) = -$3/4,000 = -$0.00075
Then the equation is something like:
y = (-$0.00075)*x + b
to find the value of b we can use one of the known points.
For example, the point (25,000 , $12) means that when x = 25,000, the price is $12
then:
$12 = (-$0.00075)*25,000 + b
$12 = -$18.75 + b
$12 + $18.75 = b
$30.75 = b
Then the equation is:
p(x) = (-$0.00075)*x + $30.75
b) We want to find the ticket price such that it maximizes the revenue.
The revenue will be equal to the price per ticket, p(x) times the total attendance, x.
Then the revenue can be written as:
r(x) = x*p(x) = x*( (-$0.00075)*x + $30.75 )
r(x) = (-$0.00075)*x^2 + $30.75*x
So we want to find the maximum revenue.
Notice that this is a quadratic equation with a negative leading coefficient, thus the maximum will be at the vertex.
Remember that for an equation like:
y = a*x^2 + bx + c
the x-value of the vertex is:
x = -b/2a
Then in our case, the x-value will be:
x = -$30.75/(2*(-$0.00075)) = 20,500
Then the revenue is maximized for x = 20,500
And the price for this x-vale is given by:
p( 20,500) = (-$0.00075)*20,500 + $30.75 = $15.375
which should be rounded to $15.38
find the area of the triangle
Answers
9514 1404 393
Answer:
108 cm²
Step-by-step explanation:
The area of a triangle is given by the formula ...
A = 1/2bh
where b represents the base length, and h represents the height--the perpendicular distance from the base to the opposite vertex. The area of this triangle is ...
A = 1/2(12 cm)(18 cm) = 108 cm²
Polynomials with odd degrees typically make a "u-shaped graph" and polynomials with even degrees typically make an "s-shaped" graph.
True
False
Answers
The statement that odd degree polynomials have a u-shaped graph and even degree polynomials have an s-shaped graph is FALSE.
What do odd degree polynomials look like on a graph?Odd degree polynomials have branches that go in opposing directions which means that they will form an s-shaped graph.
Even degree polynomials on the other hand, have graphs that go in the same direction which is why they form u-shaped graphs.
In conclusion, the above statement is false.
Find out more on polynomials at https://brainly.com/question/9696642.
Illustrate the 7th pattern of the sequence of square numbers.
Answers
1,4,9,16,25,36,49,........
7th pattern =49.....
Answer:
1, 4, 9, 16, 25, 36, 49…................the 7 the pattern is 49
Addition prop of equality
subtraction prop of quality
multiplication prop of equality
Division prop of equality
simplifying
distrib prop
Answers
2simplifying
3 Addition prop
4 simplifying
help with 1 b please. using ln.
Answers
Answer:
[tex]\displaystyle \frac{dy}{dx} = \frac{1}{(x - 2)^2\sqrt{\frac{x}{2 - x}}}[/tex]
General Formulas and Concepts:
Pre-Algebra
Equality PropertiesAlgebra I
Terms/CoefficientsFactoringExponential Rule [Root Rewrite]: [tex]\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}[/tex]Algebra II
Natural logarithms ln and Euler's number eLogarithmic Property [Exponential]: [tex]\displaystyle log(a^b) = b \cdot log(a)[/tex]Calculus
Differentiation
DerivativesDerivative NotationImplicit DifferentiationDerivative Property [Multiplied Constant]: [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]
Derivative Property [Addition/Subtraction]: [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]
Basic Power Rule:
f(x) = cxⁿf’(x) = c·nxⁿ⁻¹Derivative Rule [Quotient Rule]: [tex]\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}[/tex]
Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Step-by-step explanation:
*Note:
You can simply just use the Quotient and Chain Rule to find the derivative instead of using ln.
Step 1: Define
Identify
[tex]\displaystyle y = \sqrt{\frac{x}{2 - x}}[/tex]
Step 2: Rewrite
[Function] Exponential Rule [Root Rewrite]: [tex]\displaystyle y = \bigg( \frac{x}{2 - x} \bigg)^\bigg{\frac{1}{2}}[/tex][Equality Property] ln both sides: [tex]\displaystyle lny = ln \bigg[ \bigg( \frac{x}{2 - x} \bigg)^\bigg{\frac{1}{2}} \bigg][/tex]Logarithmic Property [Exponential]: [tex]\displaystyle lny = \frac{1}{2}ln \bigg( \frac{x}{2 - x} \bigg)[/tex]Step 3: Differentiate
Implicit Differentiation: [tex]\displaystyle \frac{dy}{dx}[lny] = \frac{dy}{dx} \bigg[ \frac{1}{2}ln \bigg( \frac{x}{2 - x} \bigg) \bigg][/tex]Logarithmic Differentiation [Derivative Rule - Chain Rule]: [tex]\displaystyle \frac{1}{y} \ \frac{dy}{dx} = \frac{1}{2} \bigg( \frac{1}{\frac{x}{2 - x}} \bigg) \frac{dy}{dx} \bigg[ \frac{x}{2 - x} \bigg][/tex]Chain Rule [Basic Power Rule]: [tex]\displaystyle \frac{1}{y} \ \frac{dy}{dx} = \frac{1}{2} \bigg( \frac{1}{\frac{x}{2 - x}} \bigg) \bigg[ \frac{2}{(x - 2)^2} \bigg][/tex]Simplify: [tex]\displaystyle \frac{1}{y} \ \frac{dy}{dx} = \frac{-1}{x(x - 2)}[/tex]Isolate [tex]\displaystyle \frac{dy}{dx}[/tex]: [tex]\displaystyle \frac{dy}{dx} = \frac{-y}{x(x - 2)}[/tex]Substitute in y [Derivative]: [tex]\displaystyle \frac{dy}{dx} = \frac{-\sqrt{\frac{x}{2 - x}}}{x(x - 2)}[/tex]Rationalize: [tex]\displaystyle \frac{dy}{dx} = \frac{-\frac{x}{2 - x}}{x(x - 2)\sqrt{\frac{x}{2 - x}}}[/tex]Rewrite: [tex]\displaystyle \frac{dy}{dx} = \frac{-x}{x(x - 2)(2 - x)\sqrt{\frac{x}{2 - x}}}[/tex]Factor: [tex]\displaystyle \frac{dy}{dx} = \frac{-x}{-x(x - 2)^2\sqrt{\frac{x}{2 - x}}}[/tex]Simplify: [tex]\displaystyle \frac{dy}{dx} = \frac{1}{(x - 2)^2\sqrt{\frac{x}{2 - x}}}[/tex]Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
Book: College Calculus 10e
Which equation is represented by the graph below?
Answers
Answer:
Hello,
Answer C
Step-by-step explanation:
Since ln(1)=0
if x=1 then y=4 ==> y=ln(x)+4
y=ln(x) is translated up for 4 units.
Please help!
What is the pattern,
Y-interception
And equation
Answers
Answer: y=1x+1
Step-by-step explanation:
y=1x+3
that should be it
We are given a weighted coin (with one side heads, one side tails), and we want to estimate the unknown probability pp that it will land heads. We flip the coin 1000 times and it happens to land heads 406 times. Give answers in decimal form, rounded to four decimal places (or more). We estimate the chance this coin will land on heads to
Answers
Answer:
0.4060
Step-by-step explanation:
To calculate the sample proportion, phat, we take the ratio of the number of preferred outcome to the total number of trials ;
Phat = number of times coin lands on head (preferred outcome), x / total number of trials (total coin flips), n
x = 406
n = 1000
Phat = x / n = 406/ 1000 = 0.4060
The estimate of the chance that this coin will land on heads is 0.406
Probability is the likelihood or chance that an event will occur.Probability = Expected outcome/Total outcomeIf a coin is flipped 1000 times, the total outcomes will 1000
If it landed on the head 406 times, the expected outcome will be 406.
Pr(the coin lands on the head) = 406/1000
Pr(the coin lands on the head) = 0.406
Hence the estimate of the chance that this coin will land on heads is 0.406
Learn more on probability here: https://brainly.com/question/14192140
Find the multiplicative inverse of: -3/7 X -4/9
Answers
Hi there!
»»————- ★ ————-««
I believe your answer is:
[tex]\frac{21}{4}[/tex]
»»————- ★ ————-««
Here’s why:
⸻⸻⸻⸻
[tex]\boxed{\text{Calculating the answer...}}\\\\---------------\\\rightarrow -\frac{3}{7} * -\frac{4}{9}\\\\\rightarrow \frac{12}{63} \\\\\rightarrow \frac{12/3}{63/3}\\\\\rightarrow\boxed{\frac{4}{21}}\\--------------\\\rightarrow \frac{4}{21}* x= 1\\\\\rightarrow (21)*\frac{4}{21}x= 1(21)\\\\\rightarrow 4x=21\\\\\rightarrow \frac{4x=21}{4}\\\\\rightarrow \boxed{x=\frac{21}{4}}[/tex]
»»————- ★ ————-««
Hope this helps you. I apologize if it’s incorrect.
if sin150=1/2 then find sin75
Answers
Answer:
0.966
Step-by-step explanation:
When typed into a calculator, sin75 = -0.3877816354
Upon converting to degrees, the full answer is 0.96592582628
If per unit variable cost of a product is Rs.8 and fixed cost is Rs 5000 and it is sold for Rs 15 per unit, profit in 1000 units is.......
a.. rs 7000
b. rs 2000
c. rs 25000
d. rs 0
Answers
Answer:
a.. rs 7000
Because 15×1000=15000 it is SP when selling 1000units in the rate of Rs 15/unit& 8×1000=8000 this is cp when buying 1000 units in the rate of Rs 8/unit.
So,by formula of profit,
Rs (15000-8000)=Rs7000
Use the information below to complete the problem: p(x)=1/x+1 and q(x)=1/x-1 Perform the operation and show that it results in another rational expression. p(x) + q(x)
Answers
Answer:
hope u will understand...if u like this answer plz mark as brainlist
Answer:
[tex]\displaystyle p(x) + q(x) = \frac{2x}{(x+1)(x-1)}[/tex]
The result is indeed another rational expression.
Step-by-step explanation:
We are given the two functions:
[tex]\displaystyle p(x) = \frac{1}{x+1}\text{ and } q(x) = \frac{1}{x-1}[/tex]
And we want to perform the operation:
[tex]\displaystyle p(x) + q(x)[/tex]
And show that the result is another rational expression.
Add:
[tex]\displaystyle = \frac{1}{x+1} + \frac{1}{x-1}[/tex]
To combine the fractions, we will need a common denominator. So, we can multiply the first fraction by (x - 1) and the second by (x + 1):
[tex]\displaystyle = \frac{1}{x+1}\left(\frac{x-1}{x-1}\right) + \frac{1}{x-1}\left(\frac{x+1}{x+1}\right)[/tex]
Simplify:
[tex]=\displaystyle \frac{x-1}{(x+1)(x-1)} + \frac{x+1}{(x+1)(x-1)}[/tex]
Add:
[tex]\displaystyle = \frac{(x-1)+(x+1)}{(x+1)(x-1)}[/tex]
Simplify. Hence:
[tex]\displaystyle p(x) + q(x) = \frac{2x}{(x+1)(x-1)}[/tex]
The result is indeed another rational expression.
Subtract the integers. 22−(−10)
Answers
Answer:
32
Step-by-step explanation:
Step 1: change 22 - ( - 10) into 22 + 10
Step 2: solve it like normal
4. Jack started packing the box shown with 1-centimeter cubes.
- cubic centimeter
Select all the true statements below.
A. Jack needs to add 2 more layers to fill the box.
B. Jack packed 16 cubes into the bottom of the box.
The box is 8 centimeters long.
The box is 3 centimeters high.
E.) The volume of the box is 32 cubic centimeters.
F. The volume of the box is 16 centimeters.
Answers
Answer:
if I am not mistakedn the answe id e
please help! 50 points!
Answers
Answer:
a) forming a bell
b) 5
c) 4.7
d) mean
is the correct answer
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