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1 year ago

Helpppppppppppppppppppppppppp!

Mathematics

2 answers:

MrRa [10]1 year ago

7 0

Answer:

x = -26

Step-by-step explanation:

$5^7^x^+^5 =125^2^x^-^7\\\\5^7^x^+^5 =5^3^(^2^x^-^7^)\\\\5^7^x^+^5 =5^6^x^-^2^1$

∴ 7x + 5 = 6x - 21

7x - 6x + 5 = -21

x = -21 - 5

x = -26

Iteru [2.4K]1 year ago

5 0

Answer:

$\boxed{ \tt{x = - 26}}$

- Please see the attached picture for full solution!:)

------------------ HappY LearninG<3 -------------

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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

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Answer:

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Step-by-step explanation:

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