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Harlamova29_29 [7]

2 years ago

15

The cost T, in dollars, of tuition and fees at Addison Technical College is $125 per credit plus an $85 registration fee. How ma

ny credits can a student take for $1710?

1 answer:

Jet001 [13]2 years ago

6 0

Answer:

8

Step-by-step explanation:

125 + 85=210.

210 * 8 = 1680.

Therefore, the answer is 8.

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Formula One race cars can reach speeds of approximately 100 meters per second. What is this speed in meters per minute?

WARRIOR [948]

Set up multiplication/division of ratios to cancel out units you don't need for the final answer and introduce those that you do.

We have m/s and we want m/min so

(100m/s)(60s/min)=6000m/min

5 0

3 years ago

What number must be added to the expression below to complete the square?

stiv31 [10]

D. 1/4

Explanation:

a perfect square is written in the form

$(x+a)^2 = x^2 + 2ax +a^2$ (1)

In our problem, we have this perfect square:

$x^2 - x + C$ (2)

with C being unknown. However, by comparing (1) and (2), we know that

$2a=-1$

So, we find

$a=-\frac{1}{2}$

and therefore, the missing term must be

$a^2 = \frac{1}{4}$

so the complete square is

$x^2 -x +\frac{1}{4}$

7 0

2 years ago

Read 2 more answers

Two sides of a triangle are 4 meters and 12 meters long . Which of the following is not a possible length of the third side?

natulia [17]

Answer:

it is D

Step-by-step explanation:

:)

7 0

1 year ago

Read 2 more answers

What is 1/5 times 10x?

grigory [225]

Answer:

2x

Step-by-step explanation:

10x/5 is the same as 1/5 times 10x.

Simplify to get 2x

8 0

3 years ago

Read 2 more answers

Let f be defined by the function f(x) = 1/(x^2+9)

riadik2000 [5.3K]

(a)

$\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}$

Substitute <em>x</em> = 3 tan(<em>t</em> ) and d<em>x</em> = 3 sec²(<em>t </em>) d<em>t</em> :

$\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt$

$=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}$

(b) The series

$\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}$

converges by comparison to the convergent <em>p</em>-series,

$\displaystyle\sum_{n=3}^\infty\frac1{n^2}$

(c) The series

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}$

converges absolutely, since

$\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}$

That is, ∑ (-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ converges absolutely because ∑ |(-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ| = ∑ (<em>n</em> ² + 9)/<em>e</em>ⁿ in turn converges by comparison to a geometric series.

5 0

2 years ago