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2 years ago

Help anyone can help me do this question 3,I will mark brainlest.

Mathematics

1 answer:

malfutka [58]2 years ago

8 0

Answer:

see explanation

Step-by-step explanation:

Calculate the slopes between pairs of the 3 points using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (- 9, 3) and (x₂, y₂ ) = (- 3, 9)

m = $\frac{9-3}{-3-(-9)}$ = $\frac{6}{-3+9}$ = $\frac{6}{6}$ = 1

Repeat with (x₁, y₁ ) = (5, 1) and (x₂, y₂ ) = (- 3, 9)

m = $\frac{9-1}{-3-5}$ = $\frac{8}{-8}$ = - 1

If lines are perpendicular then the product of their slopes = - 1 , then

1 × - 1 = -1

Thus there is a right angle between the 2 lines

Then triangle is right- angled

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Help me? idk the answer :P

Alexus [3.1K]

$\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\
-------------------------------\\\\
\left( 2^8\cdot 3^{-5}\cdot 6^0 \right)^{-2}\left( \cfrac{3^{-2}}{2^3} \right)^4\cdot 2^{28}\impliedby \textit{let's do the first group}
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$

$\bf \left( 2^8\cdot \cfrac{1}{3^5}\cdot 1 \right)^{-2}\implies \left( \cfrac{2^8}{3^5} \right)^{-2}\implies \left( \cfrac{3^5}{2^8} \right)^{2}\implies \cfrac{3^{2\cdot 5}}{2^{2\cdot 8}}\implies \boxed{\cfrac{3^{10}}{2^{16}}}\\\\
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\textit{now the second group}\qquad \left( \cfrac{3^{-2}}{2^3} \right)^4\implies \left( \cfrac{\frac{1}{3^2}}{2^3} \right)^4\implies \left( \cfrac{1}{2^3\cdot 3^2} \right)^4$

$\bf \cfrac{1^4}{2^{4\cdot 3}\cdot 3^{4\cdot 2}}\implies \boxed{\cfrac{1}{2^{12}\cdot 3^8}}\\\\
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$\bf 3^{10-8}\cdot 2^{28-28}\implies 3^2\cdot 1\implies 3^2\implies 9$

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3 years ago

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part A) if the final length is 9.19, we can set f(m) = 9.19 and solve for m

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3 years ago

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2 years ago

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2 years ago

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Help anyone can help me do this question 3,I will mark brainlest.​

Help anyone can help me do this question 3,I will mark brainlest.