3 years ago

Given that AB parallels QR, AB bisects AE, QR bisects QT

Mathematics

1 answer:

enyata [817]3 years ago

5 0

Answer:

1. Dilate ΔABE by a factor of 2/5 to make ΔA'B'E'

2. Translate A' to Q

Step-by-step explanation:

We notice the triangles have the same orientation, so no reflection or rotation is involved. The desired mapping can be accomplished by dilation and translation:

1. Dilate ΔABE by a factor of 2/5 to make ΔA'B'E'

2. Translate A' to Q

The result will be that ΔA"B"E" will lie on top of ΔQRT, as required.

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The graph below shows the solution to which system of inequalities?

solong [7]

Answer:

Step-by-step explanation:

The graph shows it clearly.

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3 years ago

A parking garage charges $5.50 to park for 4 hours and $7.75 to park for 7 hours. If the cost is a linear function of the number

Oksana_A [137]

Y=mx+b
m=slope
b=yint

two points
cost=y
(4,5.5) and (7,7.75)
slope=(y2-y1)/(x2-x1)
slope=(7.75-5.5)/(7-4)=2.25/3=0.75

y=0.75x+b
find b
(4,5.5)
x=4
y=5.5
5.5=0.75(4)+b
5.5=3+b
minus 3 fromboth sides
2.5=b

y=0.75x+2.5
3 hours
y=0.75(3)+2.5
y=2.25+2.5
y=4.75

answer is D

7 0

4 years ago

Read 2 more answers

13 is subtracted from the product of 4 and a certain number. The result is equal to the sum of 5 and the original number. Find t

alexira [117]

Answer:

The number is 6.

Step-by-step explanation:

$4x-13=x+5\\3x-13=5\\3x=18\\x=6$

8 0

3 years ago

1-234 +0.1-(-1.35)<br> What is the value

vesna_86 [32]

Answer:

-231.55

Step-by-step explanation:

4 0

4 years ago

Find an equation of the tangent line to the curve 2(x2+y2)2=25(x2−y2) (a lemniscate) at the point (−3,1). An equation of the tan

valina [46]

$2(x^2+y^2)^2=25(x^2-y^2)$

Let $y=y(x)$ , so that differentiating both sides wrt $x$ gives

$4(x^2+y^2)\left(2x+2y\dfrac{\mathrm dy}{\mathrm dx}\right)=25\left(2x-2y\dfrac{\mathrm dy}{\mathrm dx}\right)$

If $x=-3$ and $y=1$ , the above reduces to

$40\left(-6+2\dfrac{\mathrm dy}{\mathrm dx}\right)=25\left(-6-2\dfrac{\mathrm dy}{\mathrm dx}\right)\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac9{13}$

This is the slope of the tangent line, which has equation

$y-1=\dfrac9{13}(x+3)\implies\boxed{y=\dfrac{9x+40}{13}}$

7 0

3 years ago