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2 years ago

If a certian perosn has a mass of 250kg on earth, would that person's mass be on the moon, if the moon's gravity is 1.62m/s2? *

Mathematics

1 answer:

IceJOKER [234]2 years ago

6 0

Answer:

Weight On the Moon: 41.34 kilograms

Step-by-step explanation:

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Six oranges cost $1.50. how much does 15 oranges cost?<br>

kati45 [8]

Answer:

3.75 for 15 oranges

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

What equation is graphed in this figure?

Ksivusya [100]

to get the equation of any straight line, we simply need two points off of it, let's use those two points in the picture below.

$(\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{5}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{5}-\stackrel{y1}{2}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{0}}}\implies \cfrac{3}{1}\implies 3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_2=m(x-x_2) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_2}{5}=\stackrel{m}{3}(x-\stackrel{x_2}{1})$

keeping in mind that for the point-slope form, either point will do, in this case we used the second one, but the first one would have worked just the same.

7 0

1 year ago

Which expression is equivalent??? help!

dexar [7]

Answer:

The equivalent expression for the given expression $\sqrt[3]{256x^{10}y^{7} }$ is

$4x^{3} y^{2}(\sqrt[3]{4xy} )$

Step-by-step explanation:

Given:

$\sqrt[3]{256x^{10}y^{7} }$

Solution:

We will see first what is Cube rooting.

$\sqrt[3]{x^{3}} = x$

Law of Indices

$(x^{a})^{b}=x^{a\times b}\\and\\x^{a}x^{b} = x^{a+b}$

Now, applying above property we get

$\sqrt[3]{256x^{10}y^{7} }=\sqrt[3]{(4^{3}\times 4\times (x^{3})^{3}\times x\times (y^{2})^{3}\times y )} \\\\\textrm{Cube Rooting we get}\\\sqrt[3]{256x^{10}y^{7} }= 4\times x^{3}\times y^{2}(\sqrt[3]{4xy}) \\\\\sqrt[3]{256x^{10}y^{7} }= 4x^{3}y^{2}(\sqrt[3]{4xy})$

∴ The equivalent expression for the given expression $\sqrt[3]{256x^{10}y^{7} }$ is

$4x^{3} y^{2}(\sqrt[3]{4xy} )$

5 0

3 years ago

Need help with homework

motikmotik

We have two points describing the diameter of a circumference, these are:

$\begin{gathered} A=(-12,-4) \\ B=(-4,-10) \end{gathered}$

Recall that the equation for the standard form of a circle is:

$(x-h)^2+(y-k)^2=r^2$

Where (h,k) is the coordinate of the center of the circle, to find this coordinate, we find the midpoint of the diameter, that is, the midpoint between points A and B.

For this we use the following equation:

$M=(\frac{x_1+x_2_{}_{}}{2},\frac{y_1+y_2}{2})$

Now, we replace and solve:

$\begin{gathered} M=(\frac{-12+(-4)}{2},\frac{-4+(-10)}{2} \\ M=(\frac{-12-4}{2},\frac{-4-10}{2}) \\ M=(\frac{-16}{2},\frac{-14}{2}) \\ M=(-8,-7) \end{gathered}$

The center of the circle is (-8,-7), so:

$\begin{gathered} h=-8 \\ k=-7 \end{gathered}$

On the other hand, we must find the radius of the circle, remember that the radius of a circle goes from the center of the circumference to a point on its arc, for this we use the following equation: