What is the length of PR ? A. 23π B. 83π C. 43π D. 163π

The sum of three times a number and −4 is less than 17?

Pavlova-9 [17]

Answer:

3(n + 4) ≤ 10 is the equation and the solution is n ≤ -2/3 Decimal Form: -0.666667

hope this helps you

Step-by-step explanation:

7 0

3 years ago

What are the coordinates of R' after a dilation by a scale factor of 3, centered at the origin, followed by the translation (x,

emmainna [20.7K]

9514 1404 393

Answer:

C. (9, 18)

Step-by-step explanation:

Given R(3, 5), a dilation by a factor of 3 will multiply each coordinate to give ...

(9, 15)

Then the translation up by 3 will give you the image coordinates ...

(x, y) ⇒ (x, y+3)

(9, 15) ⇒ (9, 15 +3) = (9, 18)

The coordinates of R' are (9, 18), matching choice C.

_____

Comment on multiple choice answers

As it often does, here it works to "guess" the answer that has the coordinates that are most-repeated. An x-coordinate of 9 is part of 3 answer choices; a y-coordinate of 18 is the only repeated value in the answer choices. If you were to guess, an appropriate guess would be (9, 18). That happens to be correct in this case.

8 0

3 years ago

Can someone help me with this plz plz !!

saul85 [17]

There is no question

6 0

3 years ago

Read 2 more answers

Can someone help me with this question?

sesenic [268]

Answer:

One of the angles in the triangle might be 50.

AND

The length of the third side must be 11cm or smaller.

Step-by-step explanation:

-The triangle might be an equilateral triangle (having all the same sides and angles). False, since the triangle sum theorem states that all angles inside of a triangle must add up to 180, so an equilateral triangle would need to have all three angles at 60 degrees.

-One of the angles in the triangle must be 120 (false; it can be anything above 90, which is not only 120)

-The length of the third side must be 11cm or smaller. (True, Triangle Inequality Theorem)

-One of the angles in the triangle might be 50 (possibly, so very much true)

3 0

2 years ago

"find the reduction formula for the integral" sin^n(18x)

dexar [7]

Let

$I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx$
For demonstration on how to tackle this sort of problem, I'll only work through the case where $n$ is odd. We can write

$\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx$
$\implies I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$

For the remaining integral, we can integrate by parts, taking

$u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx$ $\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax$

$\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx$

For this next integral, we rewrite the integrand

$\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax$
$\implies\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx=I(n-4,a)-2I(n-2,a)+I(n,a)$

So putting everything together, we found

$I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$
$I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)$
$I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$
$\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$

$\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax$

7 0

2 years ago