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2 years ago

What letter is located at approximately

Mathematics

1 answer:

Svetradugi [14.3K]2 years ago

6 0

Answer: B

Step-by-step explanation:

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National Textile installed a new textile machine in one of its factories at a cost of $250,000. The machine is depreciated linea

kirill115 [55]

Answer:

The required expression is $y=250,000-24,000 t$

Step-by-step explanation:

Consider the provided information.

National Textile installed a new textile machine in one of its factories at a cost of $250,000.

That means for 0 year the cost of the machine is $250,000.

This can be written as: (0, 250,000)

Over 10 years with a scrap value of $10,000

That means for 10 year the cost of the machine is $10,000.

This can be written as: (10, 10,000)

Now first find the rate of change:

Rate of change=m= $\frac{Rise}{Run}=\frac{10,000-250,000}{10-0}$

Rate of change=m= $\frac{Rise}{Run}=\frac{-240,000}{10}$

Rate of change=m= $\frac{Rise}{Run}=-24,000$

Now use the point-slope form of an equation of a line with the point (0, 250,000), we can find the required expression.

$(y-250,000)=-24,000(t-0)$

$y-250,000=-24,000 t$

$y=250,000-24,000 t$

Hence, the required expression is $y=250,000-24,000 t$

6 0

3 years ago

A and B are independent events. Find P(A and B). P(A)=1/5, P(B)=3/7

Temka [501]

Answer:

Step-by-step explanation:

8 0

3 years ago

∠BCD is reflected over the y-axis. If m∠BCD = 65°, what is the m∠B'C'D'?

daser333 [38]

The answer is 35

Hope this helped heheh :)

5 0

3 years ago

Read 2 more answers

What is the explicit formula for this sequence?<br> 7, 2, -3, -8...

blsea [12.9K]

Answer:

The explicit formula of that sequence is T - 5

Step-by-step explanation:

Let T represent each term in the sequence. So now try replacing T with each term in the sequence. Like this ;

7 - 5 = 2

2 - 5 = -3

-3 - 5 = -8

hope this helps

5 0

3 years ago

The harmonic motion of a particle is given by f(t) = 2 cos(3t) + 3 sin(2t), 0 ≤ t ≤ 8. (a) When is the position function decreas

iren [92.7K]

For the last part, you have to find where $f'(t)$ attains its maximum over $0\le t\le8$ . We have

$f'(t)=-6\sin3t+6\cos2t$

so that

$f''(t)=-18\cos3t-12\sin2t$

with critical points at $t$ such that

$-18\cos3t-12\sin2t=0$

$3\cos3t+2\sin2t=0$

$3(\cos^3t-3\cos t\sin^2t)+4\sin t\cos t=0$

$\cos t(3\cos^2t-9\sin^2t+4\sin t)=0$

$\cos t(12\sin^2t-4\sin t-3)=0$

So either

$\cos t=0\implies t=\dfrac{(2n+1)\pi}2$

$12\sin^2t-4\sin t-3=0\implies\sin t=\dfrac{1\pm\sqrt{10}}6\implies t=\sin^{-1}\dfrac{1\pm\sqrt{10}}6+2n\pi$

where $n$ is any integer. We get 8 solutions over the given interval with $n=0,1,2$ from the first set of solutions, $n=0,1$ from the set of solutions where $\sin t=\dfrac{1+\sqrt{10}}6$ , and $n=1$ from the set of solutions where $\sin t=\dfrac{1-\sqrt{10}}6$ . They are approximately