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4 years ago

Yeet! I need some help! 16 points to whoever answers!

Mathematics

2 answers:

Naddik [55]4 years ago

7 0

1 should be b and 2 should be a

Gre4nikov [31]4 years ago

7 0

I think one is b. And two is a.

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Find the product of the binomials,<br> (x - 3)(x + 2) (x + 4) =

erastovalidia [21]

Answer:

Step-by-step explanation:

(x - 3)(x + 2) (x + 4) =

x^3+3x^2-10x-24

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3 years ago

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Which table shows a proportional relationship between x and y?

oksano4ka [1.4K]

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2 years ago

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Find the mass of the triangle with vertices (1; 0; 0), (0; 2; 0), (0; 1; 1) if the density function is given by (x; y; z)

Luda [366]

Answer:

$=\frac{17}{3}$

Step-by-step explanation:

Density of a function is $\rho(x,y)=x^{2} +y^{2}$

I have drawn the right angle triangle for visualization

equation for the line passing through (1,0) and (0,4) is

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= $\int_{0}^1\int_{0}^{4-4x} (x^{2}+y^{2})dydx$

= $\int_{0}^{1}(x^{2}+\frac{y^{3} }{3})dx$

= $\int_{0}^{1}(x^{2}(4-4x)+\frac{1}{3}(4-4x)^{3} )dx$

= $\int_{0}^{1}(\frac{-76}{3}x^{3} +68x^{2}-64x^{}+\frac{64}{3} )dx$

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3 years ago

How do you integrate arctan(x dx? i think that if you simplify the integral you get:?

Temka [501]

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$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\int\frac x{1+x^2}\,\mathrm dx$

For the remaining integral, setting $y=1+x^2$ gives $\dfrac{\mathrm dy}2=x\,\mathrm dx$ , so

$\displaystyle\int\frac x{1+x^2}\,\mathrm dx=\frac12\int\frac{\mathrm dy}y=\frac12\ln|y|+C=\frac12\ln(1+x^2)+C$

Putting everything together, you end up with

$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\frac12\ln(1+x^2)+C$

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3 years ago

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alexandr402 [8]

Answer:

(2-c,y)

Step-by-step explanation:

idk this is what algebra nation told me so yeah

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3 years ago