How do I solve for n?

Mathematics

1 answer:

Finger [1]2 years ago

3 0

Simple just do normal simple algebra:)

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A 15 ft. telephone pole has a wire that extends from the top of the pole to the ground. The wire and the ground form a 42° angle

Lorico [155]

Answer:

Correct choice is A

Step-by-step explanation:

Consider right triangle ABC formed by the telephone pole (side AB) and ground (side BC). In this triangle AC is the length of the wire, BC is the distance from the base of the pole to the spot where the wire touches the ground and AB=15 ft, ∠ACB=42°.

Then

$\sin 42^{\circ}=\dfrac{AB}{AC}\Rightarrow AC=\dfrac{AB}{\sin 42^{\circ}}=\dfrac{15}{\sin 42^{\circ}}\approx 22.4\ ft.$
$\tan 42^{\circ}=\dfrac{AB}{BC}\Rightarrow BC=\dfrac{AB}{\tan 42^{\circ}}=\dfrac{15}{\sin 42^{\circ}}\approx 16.7\ ft.$

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3 years ago

How do you turn 800/100 in decimal form

Delvig [45]

800 % = 8.00 or 800/100 or 8/1
800 is bigger than 100, so 8 would go before the decimal!

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2 years ago

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which container hold the most? a. cylinder with a radius of 5 in and a height of 15 in b. sphere with a radius 6in c. come with

stepan [7]

Vcylinder=hpir^2 Vsphere=(4/3)pir^3 Vcone=(1/3)hpir^2 Vcylinder=15*pi*5^2=375pi in^3 Vsphere=(4/3)*pi*6^3=288pi in^3 Vcone=(1/3)*15*pi*8^2=320pi in^3 greatest is Vcylinder at 375pi in^3 answer is A (cylinder)

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2 years ago

For the graph shown to the right, find (a) AB to the nearest tenth and (b) the coordinates of the midpoint of AB.

Shkiper50 [21]

AB =

$\sqrt{( {4 - (- 2))}^{2} + {( - 5 + 3)}^{2} } = \sqrt{40} = 6.3$
The midpoint os segment AB =

$(( \frac{ - 2 + 4}{2} ) = the \: x \: coordinate \: of \: the \: midpoint \\ \\ ( \frac{ - 3 - 5}{2})) = the \: y \: coordinate \: of \: the \: midpoint$
The midpoint is:

(1,-4)

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3 years ago

Calc help pretty please

Gre4nikov [31]

Answer:

A. 77.4

Step-by-step explanation:

Linear approximation formula

$L(x)=f(a)+f'(a)(x-a)$

Given function:

$f(x)=(x+1)^2$

$\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $y=f(u)$ and $u=g(x)$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=\dfrac{\text{d}y}{\text{d}u}\times \dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}$