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3 years ago

7

For an angle Θ with the point (5, -12) on its terminating side, what is the value of cosine

1 answer:

lorasvet [3.4K]3 years ago

3 0

Cosine = x-component/hypotenuse

cosine = 5 / √(12^2 + 5^2) = 5/13 = 0.385

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3 years ago

-5 – 3x > 2(10 + 2x) + 3<br> Solve inequality

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3 years ago

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Bill has a bike that weighs 56 pounds. Magda has a bike that weighs 52 pounds. She adds a bell and basket to her bike. That bell

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3 years ago

Solve the equation for the height?

Mars2501 [29]

Answer:

$\bold{h=\dfrac{S-2\pi r^2}{2\pi r}=\dfrac{S}{2\pi r}-r}$

Step-by-step explanation:

$\bold{S=2\pi r^2+2\pi rh}\\\\\bold{2\pi r^2+2\pi rh\ =\ S}\\\\{}\qquad -2\pi r^2\quad\ \ -2\pi r^2\\\\{}\quad \bold{2\pi rh\ =\ S-2\pi r^2}\\\\\div(2\pi r)\quad\ \div(2\pi r)\\\\\bold{h=\dfrac{S-2\pi r^2}{2\pi r}=\dfrac{S}{2\pi r}-r}$

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3 years ago

17) The coordinates of quadrilateral EFGH are E(7, 7), F(4, 8), G(2, 2), H(7, 2). Show that EF is perpendicular to FG.

strojnjashka [21]

Answer:

see explanation

Step-by-step explanation:

first we calculate the slopes of EF and FG using the slope formula

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

with (x₁, y₁ = E (7, 7 ) and F (4, 8 )

$m_{EF}$ = $\frac{8-7}{4-7}$ = $\frac{1}{-3}$ = - $\frac{1}{3}$

repeat with (x₁, y₁ ) = F (4, 8 ) and (x₂, y₂ ) = G (2, 2 )

$m_{FG}$ = $\frac{2-8}{2-4}$ = $\frac{-6}{-2}$ = 3

if 2 lines are perpendicular then the product of their slopes = - 1

here

$m_{EF}$ × $m_{FG}$ = - $\frac{1}{3}$ × 3 = - 1

then EF is perpendicular to FG

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2 years ago