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just olya [345]
3 years ago
7

How to solve for the arc CFD? (Picture)

Mathematics
1 answer:
Marat540 [252]3 years ago
3 0

Answer: 306 degrees

Explanation: CF is a semi arc making it 180 degrees. FD is 81 plus your already solved 45 degrees equaling 126. You then simply add 180 and 126 giving u your answer: 306 degrees

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If f(x) = 3x and g(x) = 1/x , what is the domain of (gºf)(x)?
Elanso [62]

Answer: 1/3x

Step-by-step explanation:

5 0
4 years ago
What is the area of the figure? Please show work?
WARRIOR [948]

Answer:

Area of ΔABC = 144 ft^2

Step-by-step explanation:

ΔABC is 45 45 90 right triangle

Ratio of leg (AC, BC) : hypo (AB)  = x : x√2

Given hypo AB = 24 so AC = BC = 12√2

Area of ABC = 1/2 bh

= 1/2 ( 12√2)( 12√2)

= 1/2(144 * 2)

= 144 ft^2

3 0
3 years ago
Which of the following measurements of capacity is the largest?
dem82 [27]

Answer:

kiloliter since 1 kiloliter is equal to 1000 liters and the rest are smaller than a kiloliter

6 0
3 years ago
Read 2 more answers
A 500 gallon tank initially contains 200 gallons of water with 5 lbs of salt dissolved in it. Water enters the tank at a rate of
Lapatulllka [165]
Until the concerns I raised in the comments are resolved, you can still set up the differential equation that gives the amount of salt within the tank over time. Call it A(t).

Then the ODE representing the change in the amount of salt over time is

\dfrac{\mathrm dA}{\mathrm dt}=\text{rate in}-\text{rate out}
\dfrac{\mathrm dA}{\mathrm dt}=\dfrac{2\text{ gal}}{1\text{ hr}}\times\dfrac{\frac15(1+\cos t)\text{ lbs}}{1\text{ gal}}-\dfrac{2\text{ gal}}{1\text{ hr}}\times\dfrac{A(t)\text{ lbs}}{500+(2-2)t}
\dfrac{\mathrm dA}{\mathrm dt}=\dfrac25(1+\cos t)-\dfrac1{250}A(t)

and this with the initial condition A(0)=5

You have

\dfrac{\mathrm dA}{\mathrm dt}+\dfrac1{250}A(t)=\dfrac25(1+\cos t)
e^{t/250}\dfrac{\mathrm dA}{\mathrm dt}+\dfrac1{250}e^{t/250}A(t)=\dfrac25e^{t/250}(1+\cos t)
\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/250}A(t)\right]=\dfrac25e^{t/250}(1+\cos t)

Integrating both sides gives

e^{t/250}A(t)=100e^{t/250}\left(1+\dfrac1{62501}\cos t+\dfrac{250}{62501}\sin t\right)+C
A(t)=100\left(1+\dfrac1{62501}\cos t+\dfrac{250}{62501}\sin t\right)+Ce^{-t/250}

Since A(0)=5, you get

5=100\left(1+\dfrac1{62501}\right)+C\implies C=-\dfrac{5937695}{62501}

so the amount of salt at any given time in the tank is

A(t)=100\left(1+\dfrac1{62501}\cos t+\dfrac{250}{62501}\sin t\right)-\dfrac{5937695}{62501}e^{-t/250}

The tank will never overflow, since the same amount of solution flows into the tank as it does out of the tank, so with the given conditions it's not possible to answer the question.

However, you can make some observations about end behavior. As t\to\infty, the exponential term vanishes and the amount of salt in the tank will oscillate between a maximum of about 100.4 lbs and a minimum of 99.6 lbs.
5 0
4 years ago
Write an equation for a translation of the parent function y = |x| to translate it 7 units left.
hoa [83]

9514 1404 393

Answer:

  B.  y = |x +7|

Step-by-step explanation:

The translations represented by the answer choices are ...

  A. up 7 units

  B. left 7 units . . . . the choice we want

  C. right 7 units

  D. down 7 units

___

In general, the effects of adding things to x and y are summarized by ...

  y = f(x -a) +b . . . . . . . . translation right 'a' units, up 'b' units

Translations left or down are accomplished by using negative values for 'a' and/or 'b.

__

We want translation 7 units left, so a=-7 and b=0

  y = |x|   ⇒   y = |x -(-7))|  or  y = |x+7|

4 0
3 years ago
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