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3 years ago

HELP ME solve for the right triangle

Mathematics

1 answer:

VikaD [51]3 years ago

3 0

Answer:

see explanation

Step-by-step explanation:

Using the cosine and tangent ratios in the right triangle

cos41° = $\frac{adjacent}{hypotenuse}$ = $\frac{VW}{VX}$ = $\frac{7}{VX}$

Multiply both sides by VX

VX × cos41° = 7 ( divide both sides by cos41° )

VX = $\frac{7}{cos41}$ ≈ 9.3

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tan41° = $\frac{opposite}{adjacent}$ = $\frac{WX}{VW}$ = $\frac{WX}{7}$

Multiply both sides by 7

7 × tan41° = WX, thus

WX ≈ 6.1

-------------------------------------------------------------------------

The sum of the 3 angles in a triangle = 180°

Subtract the sum of the given angles from 180° for ∠ X

∠ X = 180° - (90 + 41)° = 180° - 131° = 49°

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A football team has enough resources to recruit one player. The players the team is interested in consist of 7 wide receivers, 1

Alborosie

Answer:

I am not sure but i think it is 11/250

Step-by-step explanation:

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3 years ago

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

4 0

4 years ago

The box-and-whisker plot shown represents the results of a final exam. Jimmy earned an 85 on the exam. Describe how his score co

ANEK [815]

Answer:

a) False

b) False

c) False

d) True

Step-by-step explanation:

Since it is not a multiple choice question, I have answered it as a true/false question.

First, let's identify where Jimmy's score (an 85) lies in the box-and-whisker plot. It is on the upper quartile, which is also a 75% or ¾ mark compared to the scores his classmates obtained.

To the right, are those that score better than Jimmy (the scores increase to the right). Thus, to find the fraction of students that scored higher than him, we take 1 whole minus ¾, which would give us ¼.

Fraction of students that scored lower than Jimmy= ¾

(Since when we count the position of Jimmy's score, we count it with respect to the lowest score in the class.)

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3 years ago

Please help me answer this :) <br> Explain your answer <br><br> Will give brainiest

lora16 [44]

Answer:

A: Store A will cost about $1.28 less

Step-by-step explanation:

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3 years ago

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Combine like termes to simplify 4x-7+7x-9

professor190 [17]

4x-9

work shown:
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3 years ago