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Rufina [12.5K]
3 years ago
11

A driver receives -25 points for each rule violation.what integer represents the change in points after 4 rule violations?

Mathematics
1 answer:
nikklg [1K]3 years ago
4 0

-25 + -25 + -25 + -25

= 4(-25)

= -100

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Which of the following describes the image of a triangle after a dilation that has a scale factor of 5/6
Sergeeva-Olga [200]

Answer:

See example below.

Step-by-step explanation:

The image of a triangle will be 5/6 the size. It will be slightly smaller than the original. For example, if the side measurement of the triangle on one side is 12 then the image of the triangle after a dilation will be 12*5/6 = 60/6 = 10.

The new side length is 10 slightly smaller than 12.

6 0
3 years ago
How many cubes with side lengths of 1/3 does it take to fill the prism?
Ugo [173]

Answer:

180 cubes

Step-by-step explanation:

little cubes:

v=(\frac{1}{3} )^{3} =\frac{1}{27}

The prism:

v=(\frac{5}{3})(\frac{4}{3}  )(2)=\frac{40}{9} =\frac{20}{3}

the number of cubes that fill the prism:

cubes=\frac{\frac{20}{3} }{\frac{1}{27} } =\frac{(20)(27)}{(3)(1)}= \frac{540}{3} =180

Hope this helps

8 0
2 years ago
A classmate estimated square root of 397 to be about 200 explain the mistake and correct it
Arturiano [62]
They probably thought square rotting was dividing by 2. The correct answer is about 20
5 0
3 years ago
Please help and show wprk
mr Goodwill [35]
Volume of a cylinder is \pi r^2h, where r is the radius and h is the height.

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\pi r^2(3r) \\
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6 0
4 years ago
Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

I gotta go so I'll stop here, sorry.

7 0
3 years ago
Read 2 more answers
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