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2 years ago

Help me with this please

Mathematics

1 answer:

Leya [2.2K]2 years ago

6 0

Answer:

280

Step-by-step explanation:

what i did was just go step by step in what your teacher tells you to do but i am going to say this may be wrong.

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Determine whether each pair of lines is perpendicular parallel or neither

vova2212 [387]

Parallel lines mean that the slopes are the same, while perpendicular means that the slopes are negative reciprocals from each other.

With this information,
1. The slopes are the same, so PARALLEL.
2. The slopes are negative reciprocals, so PERPENDICULAR.
3. Same as the second.

7 0

2 years ago

Read 2 more answers

Carlos is shopping for a cap. He knows that his head circumference is 22 inches. What size of cap should he buy to fit his head

Paul [167]

Answer:

$7\ in$

Step-by-step explanation:

we know that

The circumference of a circle is equal to

$C=\pi D$

where

D is the diameter of the circle

In this problem we have

$C=22\ in$

substitute and solve for D

$22=(3.14156)D$

$D=22/(3.14156)=7\ in$

7 0

3 years ago

Write the following phrase as an inequality: 3 more than twice n is at most 50.

Margarita [4]

Step-by-step explanation:

3 more than twice n = 3+2n

3 more than twice n is atmost 50,

=> 3+2n <= 50 {(3+2n) is less than or equal to 50}

7 0

3 years ago

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Convert: 67100 mm = _____ m?

GrogVix [38]

Answer:

67.1m

Step-by-step explanation:

67100 divided by 1000 = 67.1m

4 0

3 years ago

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 x = 3 tan(t ) and dx = 3 sec²(t ) dt :

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

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

(b) The series

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

converges by comparison to the convergent p-series,

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

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

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 (n ² + 9)/eⁿ converges absolutely because ∑ |(-1)ⁿ (n ² + 9)/eⁿ| = ∑ (n ² + 9)/eⁿ in turn converges by comparison to a geometric series.

5 0

3 years ago