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egoroff_w [7]
2 years ago
10

Please help me figure this out

Mathematics
2 answers:
katrin2010 [14]2 years ago
3 0

Step-by-step explanation:

standard form is given by Ax × By = C

hence the answer would be

5x + 8y = 24

comment if you still have any doubts :)

gladu [14]2 years ago
3 0

Answer:

5x + 8y = 24

Step-by-step explanation:

Given :

y=-\frac{5x}{8} +3

……………………

y=-\frac{5x}{8} +3

\Longleftrightarrow 8\times y=8\times (-\frac{5x}{8} +3)

\Longleftrightarrow 8y=-5x+24

\Longleftrightarrow 5x+8y=24

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Answer:

5 hours.

Step-by-step explanation:

5 hours * 60km = 300km in 5 hours.

3 0
3 years ago
What is the greatest common denominator
Eva8 [605]
3 is the greatest common denominator
5 0
3 years ago
(5⁴)²?????? <br><br><br>A:5⁶<br>B:5⁸<br>C:20²<br>D:25⁸​
ArbitrLikvidat [17]

Answer:c 20

Step-by-step explanation:

4 0
3 years ago
Question:
aivan3 [116]

Answer:

part A) The scale factor of the sides (small to large) is 1/2

part B) Te ratio of the areas (small to large) is 1/4

part C) see the explanation

Step-by-step explanation:

Part A) Determine the scale factor of the sides (small to large).

we know that

The dilation is a non rigid transformation that produce similar figures

If two figures are similar, then the ratio of its corresponding sides is proportional

so

Let

z ----> the scale factor

\frac{CB}{C'B'}=\frac{CD}{C'D'}=\frac{BD}{B'D'}

The scale factor is equal to

z=\frac{CB}{C'B'}

substitute

z=\frac{4}{8}

simplify

z=\frac{1}{2}

Part B) What is the ratio of the areas (small to large)?

<em>Area of the small triangle</em>

A=\frac{1}{2}(2)(4)=4\ units^2

<em>Area of the large triangle</em>

A=\frac{1}{2}(4)(8)=16\ units^2

ratio of the areas (small to large)

ratio=\frac{4}{16}=\frac{1}{4}

Part C) Write a generalization about the ratio of the sides and the ratio of the areas of similar figures

In similar figures the ratio of its corresponding sides is proportional and this ratio is called the scale factor

In similar figures the ratio of its areas is equal to the scale factor squared

4 0
3 years ago
Angle 0 lies in the second quadrant and sin 0 = 3/5 cos= coy =
Elza [17]
Consider the right triangle ABC with legs AB=4, AC=3 and hypotenuse BC=5. Angle B has
 \sin B=\frac{opposite}{hypotenuse} = \frac{3}{5} and
 \cos B = \frac{adjacent}{hypotenuse} = \frac{4}{5}.

Since O lies in second quadrant \cos O\ \textless \ 0 and 
\cos O= -\cos B =- \frac{4}{5}.
Answer: \cos O=- \frac{4}{5}. 


3 0
3 years ago
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