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Whitepunk [10]
3 years ago
5

A baseball player got 52 hits one season. He got h of the hits in one game. What expression represents the number of hits he got

in the rest of the​ games?
Mathematics
1 answer:
Mars2501 [29]3 years ago
5 0

Answer:

The expression that represents the number of hits he got in the rest of the​ games is t=52-h.

Step-by-step explanation:

Given:

Number of hits in one season = 52 hits

Number of hits in one game = 'h'

We need to write expression that represents the number of hits he got in the rest of the​ games.

Solution:

Let the number of hits he got in the rest of the​ games be 't'.

Now we can say that;

The number of hits he got in the rest of the​ games is equal to Number of hits in one season minus Number of hits in one game.

framing in equation form we get;

t=52-h

Hence expression that represents the number of hits he got in the rest of the​ games is t=52-h.

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Answer:

V = 21145.01 mm³

Step-by-step explanation:

Given that:

The radius of the gumball = 30  mm

The radius of the spherical hollow-core = 28 mm

Consider the radius of the gumball to be r_2 and the radius of the spherical hollow-core to be r_1. Then;

The volume of the gumball can be determined by using the formula of a sphere.

V = \dfrac{4}{3}\pi (r_2^3 - r_1^3)

V = \dfrac{4}{3} \times \pi \times (30^3 - 28^3)

V = 21145.01 mm³

5 0
3 years ago
4(2-4x)-3x=65 Whats the answer and how did you solve it?
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Answer:

x= -3

Step-by-step explanation:

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lastly divide -19 from both sides which leaves you with x= -3

3 0
3 years ago
In 2015, the median family income in the United States was $66,650. If the 90th percentile for the 2015 median four-person famil
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Answer:

                           

Step-by-step explanation:

                 

6 0
3 years ago
Read 2 more answers
Find the surface area of the surface given by the portion of the paraboloid z=3x2+3y2 that lies inside the cylinder x2+y2=4. (hi
natta225 [31]
Parameterize the part of the paraboloid within the cylinder - I'll call it S - by

\mathbf r(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle=\left\langle u\cos v,u\sin v,3u^2\right\rangle

with 0\le u\le2 and 0\le v\le2\pi. The region's area is given by the surface integral

\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=2}\int_{v=0}^{v=2\pi}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv
=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=2}u\sqrt{1+36u^2}\,\mathrm du\,\mathrm dv
=\displaystyle2\pi\int_{u=0}^{u=2}u\sqrt{1+36u^2}\,\mathrm du

Take w=1+36u^2 so that \mathrm dw=72u\,\mathrm du, and the integral becomes

=\displaystyle\frac{2\pi}{72}\int_{w=1}^{w=145}\sqrt w\,\mathrm dw
=\displaystyle\frac\pi{36}\frac23w^{3/2}\bigg|_{w=1}^{w=145}
=\dfrac\pi{54}(145^{3/2}-1)\approx101.522
7 0
3 years ago
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