Let the height of the rectangle = h
2b + 2h = 100
2(38) + 2h = 100
2h = 100 - 76
h = 12mm
area of rectangle = base x height
= 38 x 12
= 456mm²
Answer:
456mm²
Step-by-step explanation:
using formula;
b/2( P - 2b) = A
substitute values
Find the equation of the line that
is perpendicular to y = -4x + 3
and contains the point (8, 1).
Answer:
x-4y=8
Step-by-step explanation:
y=mx+c comparing with given eq
we get slope(m1)=-4
since both are prependicular
m1×m2=-1
-4×m2=-1
m2=1÷4
eq:-y-y1=m2 (x-x1)
y-1=(1÷4)(x-8)
x-4y=4
A tortoise moves forward 15 meters in one hour. It turns around and crawls 10 meters in the
next hour. Finally, in the third hour, it turns around again and crawls 8 more meters. How
much did the tortoise walk in total in 3 hours?
Answer:
Below.
Step-by-step explanation:
15+10+8=33.
Answer:
13 meters
Step-by-step explanation:
It went 15 meters, but then it went back 10 meters.
[tex]15-10=5[/tex]
Then it went 8 more meters.
[tex]5+8=13[/tex]
Hope this helped! Please mark brainliest :)
f(x)=2^x by a factor of five
find the period of the graph shown below
Answer:
Step-by-step explanation:
The period of a trig function tells you how much space is taken up by one "up-down-up" of the graph, which is a revolution. Because half of this graph takes place in a span of 2π, then the whole graph will span 4π.
The given equation has been solved in the table. Step Statement 1 1 –7n + 11 = -10 2. -7n + 11 – 11 = -10 – 11 3 -7n = -21 4 = = =21 .In -7 -21 __7 5 n = 3 Use the table to complete each statement. In step 2, the In step 4, the property of equality was applied. property of equality was applied.
Answer:
In step 2, the subtraction property of equality was applied
In step 4, the division property of equality was applied
Step-by-step explanation:
In ΔTUV, the measure of ∠V=90°, UT = 65, VU = 56, and TV = 33. What ratio represents the cosine of ∠T?
Answer: The ratio that represents the cosine of ∠T is [tex]\frac{56}{65}[/tex]
Step-by-step explanation:
We are given:
UV = 56 units
VT = 33 units
UT = 65 units
∠V = 90°
Cosine of an angle is equal to the ratio of base and the hypotenuse of the triangle. ΔTUV is drawn in the image below.
[tex]\cos \theta=\frac{\text{base}}{\text{hypotenuse}}[/tex]
Base of the triangle is UV and the hypotenuse of the triangle is TU
Putting values in above equation, we get:
[tex]\cos \theta=\frac{UV}{TU}=\frac{56}{65}[/tex]
Hence, the ratio that represents the cosine of ∠T is [tex]\frac{56}{65}[/tex]
A and B are two similar 2D shapes
A 12cm
B 15cm
The area of the shape A is 200cm^2.
Calculate the area of shape B
Answer: [tex]312.5\ cm^2[/tex]
Step-by-step explanation:
Given
A and B are two similar shape with lengths of 12 cm and 15 cm
A has an area of [tex]200\ cm^2[/tex]
For similar figures, ratio of the square of corresponding length is equal to the ratio of the area
[tex]\Rightarrow \dfrac{200}{A_b}=\dfrac{12^2}{15^2}\\\\\Rightarrow A_b=\dfrac{15^2}{12^2}\times 200\\\\\Rightarrow A_b=312.5\ cm^2[/tex]
Plz help. How to convert this standard notation to scientific notation 549,755,813,888.
Answer:
To change a number from scientific notation to standard form, move the decimal point to the left (if the exponent of ten is a negative number), or to the right (if the exponent is positive). You should move the point as many times as the exponent indicates. Do not write the power of ten anymore
write twelve thousand twelve hundred and twelve in numbers
Answer:
12, 120,012
Step-by-step explanation:
Solve the equation and enter the value of x below. 9x + 4 + x = 54
Hello!
9x + 4 + x = 54 <=>
<=> 9x + x + 4 = 54 <=>
<=> 10x + 4 = 54 <=>
<=> 10x = 54 - 4 <=>
<=> 10x = 50 <=>
<=> x = 50 : 10 <=>
<=> x = 5 => 9 × 5 + 4 + 5 = 54
Good luck! :)
Answer:
x = 5
Step-by-step explanation:
First, combine like terms. Like terms are terms with the same variables as well as same amount of said variables:
9x + x + 4 = 54
(9x + x) + 4 = 54
10x + 4 = 54
Next, isolate the variable, x. Note the equal sign, what you do to one side, you do tot he other. Do the opposite of PEMDAS.
PEMDAS is the order of operations, and stands for:
Parenthesis
Exponents (& Roots)
Multiplication
Division
Addition
Subtraction
-
First, subtract 4 from both sides of the equation:
10x + 4 (-4) = 54 (-4)
10x = 54 - 4
10x = 50
Next, divide 10 from both sides of the equation:
(10x)/10 = (50)/10
x = 50/10 = 5
x = 5 is your answer.
~
What is the difference between calculating the area and calculating the perimeter of a rectangle?
Answer:
For perimeter you add up the side lengths to get the perimeter but for area you multiply the length times width (L x W )to get area.
Step-by-step explanation:
Maths assignment
y^2-36
Answer:
Since both terms are perfect squares, factor using the difference of squares formula,
a ^2 − b ^2 = ( a + b ) ( a − b )
where
a = y
and
b = 6
( y + 6 ) ( y − 6 )
PLEASE HELP! Which of the following ordered pairs is a solution to the given system of equations?
A. (12, 8)
B. (3, 5)
C. (-3, 3)
D. (0, 4)
please don’t use this for points.
Answer:
A.............
Step-by-step explanation:
. ..........
Answer:
C. (3,3)
Step-by-step explanation:
When These equations are both graphed the solution for these equations when they intersect is (-3,3)
if the diagonal of a square is √48 what is the area of a square
Answer:
using Pythagoras' theorem c²=a²+b²
the diagonal is the hypotenuse of one of the triangles formed
let x represent one side of the square
√48²=x²+x²
√48²=2x²
48=2x²
48/2=2x²/2
24=x²
√24=√x²
4.8989794855663561=x
~4.90
Area of the square=side x side
4.90x4.90
24.01units²
the diagram shows a square with perimeter 20cm
work out the percentage of the area shaded inside the square that is shaded
Answer:
70%
Step-by-step explanation:
Finding each side :-
4a = 20ma = 5 mFinding area of unshaded region :-
A = 1/2 * b * h A = 1/2 * 1.5 * 5 cm² A = 2.5 * 1.5 cm²A = 3.75 cm² Total unshaded region = 2*3.75 = 7.5 cm²Overall area :-
A = a²A = (5cm)²A = 25m²Finding percentage :-
% = 25 - 7.5 / 25 * 100 % = 17.5 * 4 % = 70Answer:70%
Step-by-step explanation:
A circular garden is surrounded by a circular path of 7m width.If the area of path is 770m²,find the area of the garden without path.
help me this question ⁉️
Answer:
Answer:
Radius of the circular garden
= 210 sq
=105m
Radius of the region covering the garden and the path =105m+7m
=112m
Area of the region between two concentric circles
with radius of outer circle R, and inner circle r =π(R sq−r sq)
Hence, the area of the path
=π(112sq−105 sq)= 7/22
(12544−11025)
= 33418/7
=4774m sq
HOPE THIS WILL HELP YOU MATE
A house plan Is drawn to a scale 1cm to 2m. What is the length of a window 2.5cm long on the plan?
1cm = 2m
=> 1cm = 200cm
2.5cm = 2.5 × 200cm = 500 cm = 5m
So, the length of window is 500cm or 5m.
HELP ME !
Please!
Which of the following tables represents a function?
Circle the equation of a straight line that does not intersect the curve y = x2
(1 m
y = 5
X=-3
y = 2x - 5
y= -3x + 1
Answer:
[tex] \boxed{y = 2x\: – \: 5} [/tex]
Which of these numbers had exactly tow factors?
Select your answer. 7 8 9 10
A B C D
Answer:
prime numbers only have two factors...
7 is the only prime in the list
Step-by-step explanation:
Plz help me.
I WILL GIVE BRAINLY
Answer:
p = T - a - b
Step-by-step explanation:
T = a + p + b
p = T - a - b
2
Solve the equation log, (3t+9) - log, 21 =1
Answer:
67
Step-by-step explanation:
log(3t+9)-log21 = 1
Applying, the law of logarithm,
log(3t+9)/21 = 1
converting the log into index
(3t+9)/21 = 10
solving for t
3t+9 = 21×10
3t+9 = 210
3t = 210-9
3t = 201
t = 201/3
t = 67
If K is the midpoint of JL, JK = 8x + 11 and KL = 14x – 1, find JL.
Answer:
[tex]JL=54[/tex]
Step-by-step explanation:
We are given that K is the midpoint of JL. Using this information, we want to find JL.
By the definition of midpoint, this means that:
[tex]JK=KL[/tex]
Substitute them for their equations:
[tex]8x+11=14x-1[/tex]
Solve for x. Subtract 8x from both sides:
[tex]11=6x-1[/tex]
Add 1 to both sides:
[tex]6x=12[/tex]
And divide both sides by 6. Hence:
[tex]x=2[/tex]
JL is the sum of JK and KL. Hence:
[tex]JK+KL=JL[/tex]
Since JK = KL, substitute either one for the other:
[tex]JK+(JK)=2JK=JL[/tex]
Substitute JK for its equation:
[tex]2(8x+11)=JL[/tex]
Since we know that x = 2:
[tex]2(8(2)+11)=2(16+11)=2(27)=54=JL[/tex]
Thus:
[tex]JL=54[/tex]
W=VI. Make V the subject of formula
Answer:
hope that is helpful
Step-by-step explanation:
W= VI
W= VI
I. I
V= W
I
Answer:
V = [tex]\frac{W}{I}[/tex]
Step-by-step explanation:
Given
W = VI ( isolate V by dividing both sides by I )
[tex]\frac{W}{I}[/tex] = V
Write an
equivalent expression by distributing the
"---"
sign outside the parentheses:
-(3.9d + 10)
Answer of this question
-3.9d-10which of the following are identities? check all that apply.
A. (sinx + cosx)^2= 1+sin2x
B. sin6x=2 sin3x cos3x
C. sin3x/sinxcosx = 4cosx - secx
D. sin3x-sinx/cos3x+cosx = tanx
Answer: (a), (b), (c), and (d)
Step-by-step explanation:
Check the options
[tex](a)\\\Rightarrow [\sin x+\cos x]^2=\sin ^2x+\cos ^2x+2\sin x\cos x\\\Rightarrow [\sin x+\cos x]^2=1+2\sin x\cos x\\\Rightarrow \Rightarrow [\sin x+\cos x]^2=1+\sin 2x[/tex]
[tex](b)\\\Rightarrow \sin (6x)=\sin 2(3x)\\\Rightarrow \sin 2(3x)=2\sin (3x)\cos (3x)[/tex]
[tex](c)\\\Rightarrow \dfrac{\sin 3x}{\sin x\cos x}=\dfrac{3\sin x-4\sin ^3x}{\sin x\cos x}\\\\\Rightarrow 3\sec x-4\sin ^2x\sec x\\\Rightarrow 3\sec x-4[1-\cos ^2x]\sec x\\\Rightarrow 3\sec x-4\sec x+4\cos x\\\Rightarrow 4\cos x-\sec x[/tex]
[tex](d)\\\Rightarrow \dfrac{\sin 3x-\sin x}{\cos 3x+\cos x}=\dfrac{2\cos [\frac{3x+x}{2}] \sin [\frac{3x-x}{2}]}{2\cos [\frac{3x+x}{2}]\cos [\frac{3x-x}{2}]}\\\\\Rightarrow \dfrac{2\cos 2x\sin x}{2\cos 2x\cos x}=\dfrac{\sin x}{\cos x}\\\\\Rightarrow \tan x[/tex]
Thus, all the identities are correct.
A. Not an identity
B. An identity
C. Not an identity
D. An identity
To check whether each expression is an identity, we need to verify if the equation holds true for all values of the variable x. If it is true for all values of x, then it is an identity. Let's check each option:
A. [tex]\((\sin x + \cos x)^2 = 1 + \sin 2x\)[/tex]
To check if this is an identity, let's expand the left-hand side (LHS):
[tex]\((\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x\)[/tex]
Now, we can use the trigonometric identity [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex] to simplify the LHS:
[tex]\(\sin^2 x + 2\sin x \cos x + \cos^2 x = 1 + 2\sin x \cos x\)[/tex]
The simplified LHS is not equal to the right-hand side (RHS) 1 + sin 2x since it is missing the sin 2x term. Therefore, option A is not an identity.
B. [tex]\(\sin 6x = 2 \sin 3x \cos 3x\)[/tex]
To check if this is an identity, we can use the double-angle identity for sine:[tex]\(\sin 2\theta = 2\sin \theta \cos \theta\)[/tex]
Let [tex]\(2\theta = 6x\)[/tex], which means [tex]\(\theta = 3x\):[/tex]
[tex]\(\sin 6x = 2 \sin 3x \cos 3x\)[/tex]
The equation holds true with the double-angle identity, so option B is an identity.
C. [tex]\(\frac{\sin 3x}{\sin x \cos x} = 4\cos x - \sec x\)[/tex]
To check if this is an identity, we can simplify the right-hand side (RHS) using trigonometric identities.
Recall that [tex]\(\sec x = \frac{1}{\cos x}\):[/tex]
[tex]\(4\cos x - \sec x = 4\cos x - \frac{1}{\cos x} = \frac{4\cos^2 x - 1}{\cos x}\)[/tex]
Now, using the double-angle identity for sine, [tex]\(\sin 2\theta = 2\sin \theta \cos \theta\),[/tex] let [tex]\(\theta = x\):[/tex]
[tex]\(\sin 2x = 2\sin x \cos x\)[/tex]
Multiply both sides by 2: [tex]\(2\sin x \cos x = \sin 2x\)[/tex]
Now, the left-hand side (LHS) becomes:
[tex]\(\frac{\sin 3x}{\sin x \cos x} = \frac{\sin 2x}{\sin x \cos x}\)[/tex]
Using the double-angle identity for sine again, let [tex]\(2\theta = 2x\):[/tex]
[tex]\(\frac{\sin 2x}{\sin x \cos x} = \frac{2\sin x \cos x}{\sin x \cos x} = 2\)[/tex]
So, the LHS is 2, which is not equal to the RHS [tex]\(\frac{4\cos^2 x - 1}{\cos x}\)[/tex]. Therefore, option C is not an identity.
D. [tex]\(\frac{\sin 3x - \sin x}{\cos 3x + \cos x} = \tan x\)[/tex]
To check if this is an identity, we can use the sum-to-product trigonometric identities:
[tex]\(\sin A - \sin B = 2\sin \frac{A-B}{2} \cos \frac{A+B}{2}\)\(\cos A + \cos B = 2\cos \frac{A+B}{2} \cos \frac{A-B}{2}\)[/tex]
Let A = 3x and B = x:
[tex]\(\sin 3x - \sin x = 2\sin x \cos 2x\)\(\cos 3x + \cos x = 2\cos 2x \cos x\)[/tex]
Now, we can rewrite the expression:
[tex]\(\frac{\sin 3x - \sin x}{\cos 3x + \cos x} = \frac{2\sin x \cos 2x}{2\cos 2x \cos x} = \frac{\sin x}{\cos x} = \tan x\)[/tex]
The equation holds true, so option D is an identity.
To know more about identity:
https://brainly.com/question/28974915
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the base of a right prism is an equilateral triangle each of whose sides measures 4cm.the altitude of the prism measures 5cm.Find the volume of the prism
Answer:
[tex]V=34.64\ cm^3[/tex]
Step-by-step explanation:
Given that,
The side of an equilateral prism = 4 cm
The altitude of the prism = 5 cm
We need to find the volume of the prism. The formula for the volume of a prism is as follows :
[tex]V=A\times h[/tex]
Where
A is the area of equilateral triangle, [tex]A=\dfrac{\sqrt3}{4}a^2[/tex]
So,
[tex]V=\dfrac{\sqrt3}{4}a^2\times h\\\\V=\dfrac{\sqrt3}{4}\times 4^2\times 5\\\\V=34.64\ cm^3[/tex]
So, the volume of the prism is equal to [tex]34.64\ cm^3[/tex].
I want to know the Answers
Step-by-step explanation:
this is the correct answer you wanted to know
please mark brainliest
what is the measure of angle D?
Answer:
57
Step-by-step explanation:
Hey there I need some assistance need on this problem. What do I mean by checkpoints and how am I supposed to find the y-intercept and the slope from the given values?
Slope Formula: y2 - y1 / x2 - x1
(m and slope represent the same quantity)
m = 1 - - 5 / -4 - 0
m = 1 + 5 / -4
m = 6 / -4
m = -3/2
Now that we know the slope, we can plug the slope and one of our points into slope-intercept form (y = mx + b) and solve for b. I will be using the point (-4,1).
y = -3/2x + b
1 = -3/2(-4) + b
1 = 6 + b
b = -5
In point form, the y-intercept is (0, -5).
Therefore, to get the equation all we need to do is plug in our slope and b-value to slope-intercept form.
Equation: y = -3/2 x - 5
To check the point (-6, -14) we plug it into our equation and see if the two sides are equal.
-14 = -3/2(-6) - 5
-14 = 9 - 5
-14 = 4
-14 does not equal 4, therefore the point is NOT on the line.
Hope this helps!