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3 years ago

What is an equation of the line that passes through the points (-6, -5) and (3, -5)?

Mathematics

1 answer:

IrinaK [193]3 years ago

5 0

Answer:

plz help

Step-by-step explanation:

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The weight of gemstones, such as diamonds, rubies, and emeralds, is measured in carats. A carat is a unit of measure equal to 20

Alexxx [7]

We are told in the question that 1 carat is equal to 200 mg.

The Hope Diamond is 44.5 carats. Let's multiply the weight of 1 carat by 44.5 to get the weight of the Hope Diamond in mg:

$\frac{200mg}{1 carat}* \frac{44.5 carats}{1}=8,900 mg$

The total weight of the Hope Diamond is 8,900 mg. However the question asks for the weight in grams. There are 1000 mg in 1 gram, so let's divide the known weight of the Hope Diamond in mg by 1000 to get the weight converted to grams:

$\frac{8,900mg}{1000}=8.9grams$

Now we know that the weight of the Hope Diamond is equal to 8.9 grams.

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3 years ago

Which property is illustrated by the following statement <br><br> 2x(6)=(6)2x

aleksandrvk [35]

Communicative prop of multiplication because the numbers are moving

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3 years ago

If f(x) =4x^2 and g(x) =x+1, find (f*g) (x)

kozerog [31]

(f * g)(x) = 4x^2(x + 1)
4x^3 + 4x^2

5 0

3 years ago

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The solution to a system of three equations in three variables is__one point.

Firdavs [7]

The answer for this is B because

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2 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

3 years ago