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2 years ago

What is the result when 8x^3+6x^2+15x+7 divides by 2x + 1

Mathematics

1 answer:

sveticcg [70]2 years ago

3 0

Answer:

Step-by-step explanation:

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The total of 36 78 198 475 and 620

Alina [70]

The total of,
36
78
198
475
+620
-------
=1407

Hope this helps!=)

8 0

3 years ago

Read 2 more answers

Somebody help me?! Finding an angle measure given a triangle and parallel lines.

balandron [24]

Answer:

x=78

Step-by-step explanation:

the sum of all angles in a triangle=180

180-59=129

y>x

70+59=129

180-129=51

129-51=78

x=78

8 0

3 years ago

Your friend says that 14.5% is equivilent to 14.5

SCORPION-xisa [38]

14.5% = 0.145 = 145/1000

14.5 = 14500/1000

<span>So 14.5% is a hundredth the size of 14.5, so they are not equal and the friend is incorrect.</span>

4 0

3 years ago

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Solve the equation 58 - 10x ≤ 20 + 9x

drek231 [11]

Answer:

x≥2

Step-by-step explanation:

First, write out the equation as you have it:

$58-10x\leq20+9x$

Then, add $10x$ to both sides:

$58-10x+10x\leq20+9x+10x\\58\leq20+19x$

Next, subtract $20$ from both sides:

$58-20\leq20-20+19x\\38\leq19x$

Finally, divide both sides by $19$ :

$\frac{38}{19}\leq\frac{19x}{19}\\2\leq x$

$x\geq 2$

Therefore: the answer to this inequality/equation is: x≥2

6 0

3 years ago

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is

$\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6$

4 0

2 years ago