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3 years ago

14

Please help I need this today :(

1 answer:

yuradex [85]3 years ago

5 0

Step-by-step explanation:

Use COS method (Cos Angle = adjacent/hypotenuse)

given adjacent = 14 ft

hypotenuse = 67 ft

$\cos(x) = \frac{adjacent}{hypotenuse} \\ \cos(x) = \frac{14}{67} \\ x = {cos}^{ - 1} ( \frac{14}{67} ) \\ = 78 \: deg(nearest \: degree)$

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NEED HELP PLZ HELP ME ….

Katyanochek1 [597]

Answer:

R= -0.95

Step-by-step explanation:

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2 years ago

Plzzzzzzzzzzzzzzzzzzzzzzzzzzzzz help

nadezda [96]

<h2> $\large\mathfrak{{\pmb{\underline{\red{Given}}{\red{:}}}}}$ </h2>

∠RSU and ∠USV are complementary angles
∠RSU = (4x – 15)°
∠USV = 20°

<h2> $\large\mathfrak{{\pmb{\underline{\color{red}{To~find}}{\color{red}{:}}}}}$ </h2>

The value of x

<h2> $\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}$ </h2>

∠RSU and ∠USV are complementary angles

$\sf\implies{ \angle RSU + \angle USV = 90 \degree }$

$\sf\implies{ (4x-15) + 20 = 90 }$

$\sf\implies{ 4x-15+20=90}$

$\sf\implies{4x+5=90}$

$\sf\implies{4x=90-5}$

$\sf\implies{4x=85}$

$\sf\implies{x={\dfrac{85}{4}}}$

$\sf\implies{ {\orange{\underline{\boxed{\sf{\pmb{x=21.25}}}}}}}$

<h2> $\large\mathfrak{\therefore{\pmb{\underline{\color{gold}{Required~answer}}{\color{gold}{:}}}}}$ </h2>

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5 0

3 years ago

An arched drainage culvert can be modelled by the function:

Zanzabum

Substitute 136.5 for y.

$136.5=\frac{x(191-x)}{60}$

$8190=x(191-x)$

$8190=191x-x^{2}$

$x^{2} -191x+8190=0$

Compare this equation with $ax^{2} +bx+c=0$

a = 1, b = - 191, c = 8190

$x=\frac{-b+\sqrt{b^{2}-4ac } }{2}$ or

$x=\frac{-b-\sqrt{b^{2}-4ac } }{2}$

$x=\frac{191+\sqrt{191^{2}-4(1)(8190) } }{2}$ or

$x=\frac{191-\sqrt{191^{2}-4(1)(8190) } }{2}$

$x=\frac{191+\sqrt{36481-32760 } }{2}$ or

$x=\frac{191-\sqrt{36481-32760 } }{2}$

$x=\frac{191+\sqrt{3721 } }{2}$ or

$x=\frac{191-\sqrt{3721 } }{2}$

$x=\frac{191+61}{2}$ or

$x=\frac{191-61}{2}$

$x=\frac{252}{2}$ or

$x=\frac{130}{2}$

x = 126 or x = 65

Hence, the points on the curve are (65, 136.5) and (126, 136.5).

The width of the air space is the distance between these points.

Width = $\sqrt{(x_{2}- x_{1})^{2} +( y_{2}- y_{1}) ^{2}$

= $\sqrt{(126- 65)^{2} +(136.5-136.5) ^{2}$

= 126 - 65

= 61

Hence, width of the air space is 61m.

5 0

3 years ago

16 ÷ 22 + (6 x 2) - 8

stepladder [879]

I did the PEMDAS method (P= parentheses, E= exponents, M= multiplication, D = Division, A= addition, S = subtraction.)

First I did what’s inside the parentheses, (6x2) and got 12.

Then I divided 16 by 22 and received 0.72.

I added 0.72 with 12 and the result was 12.72.

Last step was to subtract 12.72 - 8 and that equals 4.72

3 0

3 years ago

Read 2 more answers

What is the measure of ∠ABC?

____ [38]

Answer:

not sure

Step-by-step explanation:

I'm not sure how to do this guys I am very sorry for the inconvenience

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2 years ago

Read 2 more answers