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Nataliya [291]
4 years ago
12

How do you solve a trigonometry problem if you have one side length of 12 and two angles one being 90° and the other being 54°?

Mathematics
1 answer:
Deffense [45]4 years ago
8 0
Hello Jade,

To solve the problem, you would need to know which side has a measurement of 12. Do you have a picture of the problem?

I would be happy to help,
MrEQ
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An airplane takes off from the ground and reaches a height of 500 feet after flying 2 miles. given the formula h = d tan θ, wher
zloy xaker [14]

Height covered by the plane=h= 500 feet

Distance covered by the plane along the ground=d= 2 mile

Now, convert the mile into feet,

1 mile = 5280 feet

2 mile = 10560 feet

Now, use the formula

h= d tanθ

tanθ= h/d

tanθ= 500/10560

tanθ= 25/528

Take the tan⁻¹ on both side,

θ= tan⁻¹(25/528)

θ =2.71°



5 0
3 years ago
The Watson family and the Thompson family each used their sprinklers last summer. The Watson family's sprinkler was used for hou
o-na [289]

This question s incomplete, the complete question is;

The Watson family and the Thompson family each used their sprinklers last summer. The Watson family's sprinkler was used for 15 hours. The Thompson family's sprinkler was used for 30 hours.

There was a combined total output of 1050 of water. What was the water output rate for each sprinkler if the sum of the two rates was 55L per hour  

Answer:

The Watson family sprinkler is 40 L/hr while Thompson family sprinkler is 15 L/hr

Step-by-step explanation:

Given the data in the question;

let water p rate for Watson family and the Thompson family sprinklers be represented by x and y respectively

so

x + y = 55  ----------------equ1

x = 55 - y ------------------qu2

also

15x + 30y = 1050

x + 2y = 70 --------------equ3

input equ2 into equ3

(55 - y) + 2y = 70

- y + 2y = 70 - 55

y = 15

input value of y into equ1

x + 15 = 55

x = 55 - 15

x = 40

Therefore, The Watson family sprinkler is 40 L/hr while Thompson family sprinkler is 15 L/hr

8 0
3 years ago
Match each vector operation with its resultant vector expressed as a linear combination of the unit vectors i and j.
Cloud [144]

Answer:

3u - 2v + w = 69i + 19j.

8u - 6v = 184i + 60j.

7v - 4w = -128i + 62j.

u - 5w = -9i + 37j.

Step-by-step explanation:

Note that there are multiple ways to denote a vector. For example, vector u can be written either in bold typeface "u" or with an arrow above it \vec{u}. This explanation uses both representations.

\displaystyle \vec{u} = \langle 11, 12\rangle =\left(\begin{array}{c}11 \\12\end{array}\right).

\displaystyle \vec{v} = \langle -16, 6\rangle= \left(\begin{array}{c}-16 \\6\end{array}\right).

\displaystyle \vec{w} = \langle 4, -5\rangle=\left(\begin{array}{c}4 \\-5\end{array}\right).

There are two components in each of the three vectors. For example, in vector u, the first component is 11 and the second is 12. When multiplying a vector with a constant, multiply each component by the constant. For example,

3\;\vec{v} = 3\;\left(\begin{array}{c}11 \\12\end{array}\right) = \left(\begin{array}{c}3\times 11 \\3 \times 12\end{array}\right) = \left(\begin{array}{c}33 \\36\end{array}\right).

So is the case when the constant is negative:

-2\;\vec{v} = (-2)\; \left(\begin{array}{c}-16 \\6\end{array}\right) =\left(\begin{array}{c}(-2) \times (-16) \\(-2)\times(-6)\end{array}\right) = \left(\begin{array}{c}32 \\12\end{array}\right).

When adding two vectors, add the corresponding components (this phrase comes from Wolfram Mathworld) of each vector. In other words, add the number on the same row to each other. For example, when adding 3u to (-2)v,

3\;\vec{u} + (-2)\;\vec{v} = \left(\begin{array}{c}33 \\36\end{array}\right) + \left(\begin{array}{c}32 \\12\end{array}\right) = \left(\begin{array}{c}33 + 32 \\36+12\end{array}\right) = \left(\begin{array}{c}65\\48\end{array}\right).

Apply the two rules for the four vector operations.

<h3>1.</h3>

\displaystyle \begin{aligned}3\;\vec{u} - 2\;\vec{v} + \vec{w} &= 3\;\left(\begin{array}{c}11 \\12\end{array}\right) + (-2)\;\left(\begin{array}{c}-16 \\6\end{array}\right) + \left(\begin{array}{c}4 \\-5\end{array}\right)\\&= \left(\begin{array}{c}3\times 11 + (-2)\times (-16) + 4\\ 3\times 12 + (-2)\times 6 + (-5) \end{array}\right)\\&=\left(\begin{array}{c}69\\19\end{array}\right) = \langle 69, 19\rangle\end{aligned}

Rewrite this vector as a linear combination of two unit vectors. The first component 69 will be the coefficient in front of the first unit vector, i. The second component 19 will be the coefficient in front of the second unit vector, j.

\displaystyle \left(\begin{array}{c}69\\19\end{array}\right) = \langle 69, 19\rangle = 69\;\vec{i} + 19\;\vec{j}.

<h3>2.</h3>

\displaystyle \begin{aligned}8\;\vec{u} - 6\;\vec{v} &= 8\;\left(\begin{array}{c}11\\12\end{array}\right) + (-6) \;\left(\begin{array}{c}-16\\6\end{array}\right)\\&=\left(\begin{array}{c}88+96\\96 - 36\end{array}\right)\\&= \left(\begin{array}{c}184\\60\end{array}\right)= \langle 184, 60\rangle\\&=184\;\vec{i} + 60\;\vec{j} \end{aligned}.

<h3>3.</h3>

\displaystyle \begin{aligned}7\;\vec{v} - 4\;\vec{w} &= 7\;\left(\begin{array}{c}-16\\6\end{array}\right) + (-4) \;\left(\begin{array}{c}4\\-5\end{array}\right)\\&=\left(\begin{array}{c}-112 - 16\\42+20\end{array}\right)\\&= \left(\begin{array}{c}-128\\62\end{array}\right)= \langle -128, 62\rangle\\&=-128\;\vec{i} + 62\;\vec{j} \end{aligned}.

<h3>4.</h3>

\displaystyle \begin{aligned}\;\vec{u} - 5\;\vec{w} &= \left(\begin{array}{c}11\\12\end{array}\right) + (-5) \;\left(\begin{array}{c}4\\-5\end{array}\right)\\&=\left(\begin{array}{c}11-20\\12+25\end{array}\right)\\&= \left(\begin{array}{c}-9\\37\end{array}\right)= \langle -9, 37\rangle\\&=-9\;\vec{i} + 37\;\vec{j} \end{aligned}.

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Find the area of the figure. (Sides meet at right angles.)
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The area is 70in^2 because you find the area of the square without the missing piece then subtract area of missing piece from total
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How do I find the surface area of a cone
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Find circumference and height of cone and multiply
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