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3 years ago

Need help with math quarterly please

Mathematics

2 answers:

Readme [11.4K]3 years ago

7 0

Answer:

c=40

Step-by-step explanation:

a + c=65 equation 1

7a+4c=335 equation 2

c=65-a isolate c from equation 1

7a+4(65-a)=335

7a+260-4a=335

3a=75

a=25

solve for c

a+c=65

25+c=65

c=40

check answer:

7a+4c=335

7(25)+4(40)=335

175+160=335

335=335

lesya [120]3 years ago

4 0

Put a more clearer picture can’t see men

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2 years ago

Jayden and Sheridan both tried to find the missing side of the right triangle. A right triangle is shown. One leg is labeled as

stellarik [79]

Answer:

Sheridan's Work is correct

Step-by-step explanation:

we know that

The lengths side of a right triangle must satisfy the Pythagoras Theorem

$c^{2}=a^{2}+b^{2}$

where

a and b are the legs

c is the hypotenuse (the greater side)

In this problem

Let

$a=7\ cm\\c=13\ cm$

substitute

$13^{2}=7^{2}+b^{2}$

Solve for b

$169=49+b^{2}$

$b^{2}=169-49$

$b^{2}=120$

$b=\sqrt{120}\ cm$

$b=10.95\ cm$

we have that

<em>Jayden's Work</em>

$a^{2}+b^{2}=c^{2}$

$a=7\ cm\\b=13\ cm$

substitute and solve for c

$7^{2}+13^{2}=c^{2}$

$49+169=c^{2}$

$218=c^{2}$

$c=\sqrt{218}\ cm$

$c=14.76\ cm$

Jayden's Work is incorrect, because the missing side is not the hypotenuse of the right triangle

<em>Sheridan's Work</em>

$a^{2}+b^{2}=c^{2}$

$a=7\ cm\\c=13\ cm$

substitute

$7^{2}+b^{2}=13^{2}$

Solve for b

$49+b^{2}=169$

$b^{2}=169-49$

$b^{2}=120$

$b=\sqrt{120}\ cm$

$b=10.95\ cm$

therefore

Sheridan's Work is correct

6 0

3 years ago

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Step-by-step explanation:

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