All categories

2 years ago

Please clink on the link. Don’t give me fake answers or links or you will be reported.

Mathematics

2 answers:

Andreyy892 years ago

6 0

D is the correct answer

kow [346]2 years ago

6 0

Answer:

it's B

Step-by-step explanation:

(x + 5)^2 + (y - 2)^2 = 81

A. is not A

(1 + 5)^2 + (-4 - 2)^2

6^2 + -6^2

36 + (-36) = 0

B. is B

(-5 + 5)^2 + (11 - 2)^2

0^2 + 9^2

0 + 81 = 81

C. is not C

(-5 + 5)^2 + (2 - 2)^2

0^2 + 0^2

0 + 0 = 0

D. is not D

(-3 + 5)^2 + (-8 - 2)^2

2^2 + -10^2

4 + -100 = -98

You might be interested in

Ben is going to rotate a figure in Quadrant IV 90 counterclockwise. In which Quadrant should be image be

fomenos

The answer would be quadrant 1.

6 0

2 years ago

What is a nativist ? Help please

11111nata11111 [884]

Answer:

Nativism is the political policy of promoting the interests of native inhabitants against those of immigrants, including the support of immigration-restriction measures

Step-by-step explanation:

Hope this helps.

Please mark brainliest!

Would be very much appreciated!

7 0

3 years ago

Read 2 more answers

I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

4 0

2 years ago

Read 2 more answers

Dan sold 40 concert tickets in 5 days. Each day he sold 3 tickets more than the previous day the number of tickets he sold on th

adoni [48]

56
step my step:
multiply add divide

5 0

2 years ago

Apply The Remainder Theorem, Fundamental Theorem, Rational Root Theorem, Descartes Rule, and Factor Theorem to find the remainde

Over [174]

9514 1404 393

Answer:

possible rational roots: ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12}

actual roots: -1, (2 ±4i√2)/3

no turning points; no local extrema

end behavior is same-sign as x-value end-behavior

Step-by-step explanation:

The Fundamental Theorem tells us this 3rd-degree polynomial will have 3 roots.

The Rational Root Theorem tells us any rational roots will be of the form ...

±{factor of 12}/{factor of 3} = ±{1, 2, 3, 4, 6, 12}/{1, 3}

= ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12} . . . possible rational roots

Descartes' Rule of Signs tells us the two sign changes mean there will be 0 or 2 positive real roots. Changing signs on the odd-degree terms makes the sign-change count go to 1, so we know there is one negative real root.

The y-intercept is 12. The sum of all coefficients is 22, so f(1) > f(0) and there are no positive real roots in the interval [0, 1]. Synthetic division by x-1 shows the remainder is 22 (which we knew) and all the quotient coefficients are all positive. This means x=0 is an upper bound on the real roots.

The sum of odd-degree coefficients is 3+8=11, equal to the sum of even-degree coefficients, -1+12=11. This means that -1 is a real root. Synthetic division by x+1 shows the remainder is zero (which we knew) and the quotient coefficients alternate signs. This means x=-1 is a lower bound on real roots. The quotient of 3x^2 -4x +12 is a quadratic factor of f(x):

f(x) = (x +1)(3x^2 -4x +12)

The complex roots of the quadratic can be found using the quadratic formula:

x = (-(-4) ±√((-4)^2 -4(3)(12)))/(2(3)) = (4 ± √-128)/6

x = (2 ± 4i√2)/3 . . . . complex roots

The graph in the third attachment (red) shows there are no turning points, hence no relative extrema. The end behavior, as for any odd-degree polynomial with a positive leading coefficient, is down to the left and up to the right.

4 0

2 years ago