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3 years ago

Solve and graph inequality

Mathematics

1 answer:

alina1380 [7]3 years ago

4 0

Her sister could be about four or five

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perform the steps for dividing the two polynomials using synthetic division. state the quotient, q(x), and the remainder, r(x).

Ad libitum [116K]

Answer:

This particular equation cannot be solved using synthetic division as you can only use synthetic division when divided by a linear factor.

Step-by-step explanation:

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3 years ago

Ratio equivalents to 1:2

nata0808 [166]

2:4 3:6 4:8

Any ratio equals 1:2 as long as you can divide the first number by itself to get 1 and the second number with the 1st number.
Example
2:4
Divide the first number by 2 and you get 1
Divide the second number by 2 and you get 2
So the ratio equals 1:2

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3 years ago

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What is the slope of the line graphed above

erma4kov [3.2K]

Answer:
B

Explanation:
Slope = rise/run = -2/2 = -1

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3 years ago

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Solve for x. <br><br> a²x+(a-8)=(a+8)x

Sergio [31]

$a^2x+(a-8)=(a+8)x\\\\(a+8)x=a^2x+(a-8)\ \ \ \ |subtract\ a^2x\ from\ both\ sides\\\\(a+8)x-a^2x=a-8\\\\(a+8-a^2)x=a-8\ \ \ \ |divide\ both\ sides\ by\ (a+8-a^2)\\\\\huge\boxed{x=\frac{a-8}{a+8-a^2}}\to\huge\boxed{x=\frac{8-x}{a^2-a-8}}$

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3 years ago

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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

3 years ago