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3 years ago

7

if the input value is negative, is the output value of f(z)=-4z+12 always positive or always negative?

1 answer:

expeople1 [14]3 years ago

6 0

If the input value is negative, the output value is always positive. We know this because the negative sign before the 4z will always reverse the negativity of the number that is plugged in.

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Once again co sines . Please help me and include explanation with a clear answer

vladimir1956 [14]

Using the law os cosines formula b^2 = a^2 + c^2 - 2*a*c*cos(B)

a = 17, b = 8, c = 16

8^2 = 17^2 + 16^2 - 2*17*16* cos(B)

64 = 289 + 256 - 544 * cos(B)

544*cos(B) = 289 + 256 - 64

544 * cos(B) = 481

cos (B) = 481/544

B = arccos(481/544)

B = 27.8 degrees

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3 years ago

Please help pre Algebra question

qaws [65]

Answer:

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-3=No

2=Yes

0=No

Step-by-step explanation:

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2 years ago

Luna mixes 3/4 cup of orange juice with 3/8 cup of cranberry juice. She gives 5/8 cup of juice to mags. How much is left in luna

Anuta_ua [19.1K]

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Step-by-step explanation:

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3 years ago

Can you help me in this question?

Xelga [282]

Answer:

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Step-by-step explanation:

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2 years ago

The boundary of a lamina consists of the semicircles y = 1 − x2 and y = 16 − x2 together with the portions of the x-axis that jo

oksano4ka [1.4K]

Answer:

Required center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

Step-by-step explanation:

Given semcircles are,

$y=\sqrt{1-x^2}, y=\sqrt{16-x^2}$ whose radious are 1 and 4 respectively.

To find center of mass, $(\bar{x},\bar{y})$ , let density at any point is $\rho$ and distance from the origin is r be such that,

$\rho=\frac{k}{r}$ where k is a constant.

Mass of the lamina=m= $\int\int_{D}\rho dA$ where A is the total region and D is curves.

then,

$m=\int\int_{D}\rho dA=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}rdrd\theta=k\int_{}^{}(4-1)d\theta=3\pi k$

Now, x-coordinate of center of mass is $\bar{y}=\frac{M_x}{m}$ . in polar coordinate $y=r\sin\theta$

$\therefore M_x=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\sin\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\sin\thetadrd\theta$

$=3k\int_{0}^{\pi}\sin\theta d\theta$

$=3k\big[-\cos\theta\big]_{0}^{\pi}$

$=3k\big[-\cos\pi+\cos 0\big]$

$=6k$

Then, $\bar{y}=\frac{M_x}{m}=\frac{2}{\pi}$

y-coordinate of center of mass is $\bar{x}=\frac{M_y}{m}$ . in polar coordinate $x=r\cos\theta$

$\therefore M_y=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\cos\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\cos\theta drd\theta$

$=3k\int_{0}^{\pi}\cos\theta d\theta$

$=3k\big[\sin\theta\big]_{0}^{\pi}$

$=3k\big[\sin\pi-\sin 0\big]$

$=0$

Then, $\bar{x}=\frac{M_y}{m}=0$

Hence center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

3 0

3 years ago