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3 years ago

I can't get the answer!!!! >:(

Mathematics

1 answer:

Rasek [7]3 years ago

7 0

Answer:

The scale factor of the smaller rectangle to the larger rectangle is 3/2

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional and this ratio is called the scale factor

In this problem

To find out the scale factor of the smaller rectangle to the larger rectangle. set up a proportion

$\frac{9}{6}=\frac{3}{2}$ ----> ratio of corresponding heights

$\frac{12}{8}=\frac{3}{2}$ ----> ratio of corresponding bases

therefore

The scale factor of the smaller rectangle to the larger rectangle is 3/2

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Four classmates went bowling. Each paid $2.00 to rent shoes and bowled g games, paying $3.50 per player for each game. Which e

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Answer:

its A

Step-by-step explanation:

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3 years ago

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You and three friends own a paint shop. The shop's profit was $536 in the first month. You always divide the profits equally. In

ludmilkaskok [199]

Hello! I can help you with this! So, you and 3 friends are 4 people. You split $536. To find the amount each of you got, Do division. 536/4 is 134. Each friend got $134 in the first month. You made 224 in 2 months. Let’s subtract to see how much more each of them got. 224 - 134 is 90. Now, multiply by 4 to find out how much profit was made. 90 * 4 is 360. The shop made $360 in the second month.

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3 years ago

In a quadrilateral ABCD, angle A + angle D = 90°. Prove that AC^2 + BD^2 = BC^2 + AD^2

olga2289 [7]

Answer:

$AD^{2}+BC^{2}=AE^{2}+ED^{2}+BE^{2}+EC^{2}$

$AC^{2}+BD^{2}=AE^{2}+EC^{2}+BE^{2}+ED^{2}$

Hence prove.

$AC^{2}+BD^{2}=BC^{2}+AD^{2}$

Step-by-step explanation:

Given:

∠A + ∠D = 90°

We are prove to that

$AC^{2}+BD^{2}=BC^{2}+AD^{2}$

Solution:

See required figure in attached file.

We know sum of the all angles of a triangle is 180°.

So, in triangle AED.

∠A + ∠E + ∠D = 180°

∠E + (∠A + ∠D) = 180°

Now, we substitute ∠A + ∠D = 90° in above equation.

∠E + 90° = 180°

∠E = 180° - 90°

∠E = 90°

Using Pythagoras Theorem for triangle ADE and triangle BEC.

$AD^{2}=AE^{2}+ED^{2}$

$BC^{2}=BE^{2}+EC^{2}$

Now, we add both above equations.

$AD^{2}+BC^{2}=AE^{2}+ED^{2}+BE^{2}+EC^{2}$ --------(1)

Similarly, Using Pythagoras Theorem for triangle AEC and triangle BED.

$AC^{2}=AE^{2}+EC^{2}$

$BD^{2}=BE^{2}+ED^{2}$

Now, we add both above equations.

$AC^{2}+BD^{2}=AE^{2}+EC^{2}+BE^{2}+ED^{2}$ --------(2)

We get From equation 1 and equation 2.

$AC^{2}+BD^{2}=BC^{2}+AD^{2}$

Hence prove,

$AC^{2}+BD^{2}=BC^{2}+AD^{2}$

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3 years ago

A quadrilateral has vertices at A(–5, 5), B(1, 8), C(4, 2), D(–2, –2). Use slope to determine if the quadrilateral is a rectangl

lianna [129]

With the properties of rectangle in mind we must perform two verification. Verify to see if opposite sides are parallel and adjacent sides are perpendicular.

We need to determine the slope of each side, using the formula,

$m=\frac{y_2-y_1}{x_2-x_1 }$

Slope of AB

$m_{AB}=\frac{8-5}{1--5}$

$\Rightarrow m_{AB}=\frac{8-5}{1+5}$

$\Rightarrow m_{AB}=\frac{3}{6}$

$\Rightarrow m_{AB}=\frac{1}{2}$

Slope of BC

$m_{BC}=\frac{2-8}{4-1}$

$\Rightarrow m_{BC}=\frac{-6}{3}$

$\Rightarrow m_{BC}=-2$

Slope of CD

$m_{CD}=\frac{-2-2}{-2-4}$

$\Rightarrow m_{CD}=\frac{-2-2}{-2-4}$

$\Rightarrow m_{CD}=\frac{-4}{-6}$

$\Rightarrow m_{CD}=\frac{2}{3}$

Slope of AD

$m_{AD}=\frac{-2-5}{-2--5}$

$\Rightarrow m_{AD}=\frac{-2-5}{-2+5}$

$\Rightarrow m_{AD}=\frac{-7}{3}$

Verify Parallel sides

If the quadrilateral is a rectangle, then opposite sides should have the same slope. But

$m_{AD} = \frac{-7}{3} \neq m_{BC}=-2$

$m_{AB}=\frac{1}{2} \neq m_{CD}=\frac{2}{3}$

Verify Perpendicularity

And also the product of slopes of all sides with a common vertex should be -1. But

$m_{AB} \times m_{BC}=\frac{1}{2} \times -2=-1$

$m_{AB} \times m_{AD}=\frac{1}{2} \times -\frac{7}{3} \neq -1$

$\Rightarrow m_{CD} \times m_{AD}=\frac{2}{3} \times -\frac{7}{3} \neq -1$

$\Rightarrow m_{CD} \times m_{BC}=\frac{2}{3} \times -2 \neq -1$

Since the quadrilateral fails to satisfy all these conditions, the quadrilateral is not a rectangle

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3 years ago

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gladu [14]

1/9 because 1/9 is equivalent to .1 repeating, and since it repeats, it would continue with the 1 for infinite decimal places, thus making 11.11111% because you move the decimal over 2 places to make the fraction turn into a decimal.

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06/12/2017

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3 years ago