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3 years ago

Is it right ????????

Mathematics

1 answer:

Korolek [52]3 years ago

6 0

Answer:

yeah

Step-by-step explanation:

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8(1 - 3k) + 1 = -25 - 7k A) {3} B) {-11} C) {1} D) {2} Show your work

damaskus [11]

Answer:

Step-by-step explanation:

8 - 24k + 1 = -25 - 7k

-24k + 9 = -25 -7k

-17k + 9 = -25

-17k = -34

k = 2

answer is D

6 0

3 years ago

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This make no sense to me can someone help and explain it please

slava [35]

Answer:

x + 1

Step-by-step explanation:

They want you to write a generic expression to represent the total (adding) between a number (which they want you to write as 'x') and 1.

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3 years ago

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Angles 1 and 5 are what type of angle pair?

prohojiy [21]

Answer:

I think they are corresponding angles

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3 years ago

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Use the Green's Theorem to calculate the work done by the field [ F (x, y) = -3y^5 i + 5y^2x^3 j ] to move a particle along the

lianna [129]

$\displaystyle\int_C\mathbf F\cdot\mathrm d\mathbf r=\iint_R\left(\frac{\partial(5y^2x^3)}{\partial x}-\frac{\partial(-3y^5)}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

$\dfrac{\partial(5y^2x^3)}{\partial x}=15y^2x^2$
$\dfrac{\partial(-3y^5)}{\partial y}=-15y^2$

$=\displaystyle\int_{x=-2}^{x=2}\int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}}15y^2(x^2+1)\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates, the integral is equivalent to

$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=2}15r^3\sin^2\theta(r^2\cos^2\theta+1)\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle15\int_0^{2\pi}\int_0^2\left(\frac{r^5}4\sin^22\theta+r^3\sin^2\theta\right)\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\frac{15}4\left(\int_0^2r^5\,\mathrm dr\right)\left(\int_0^{2\pi}\sin^22\theta\,\mathrm d\theta\right)+15\left(\int_0^2r^3\,\mathrm dr\right)\left(\int_0^{2\pi}\sin^2\theta\,\mathrm d\theta\right)$
$=100\pi$

6 0

3 years ago

If (-1/3)x+7=4, what is the value for (1/3)x+3? A. 3 B. 6 C. 9 D. 12

Ahat [919]

Answer:

Step-by-step explanation:

there your answer

4 0

4 years ago