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Vesna [10]
3 years ago
11

The volume of a basketball is 448.92 cubic inches, and the volume of a baseball is 12.77 cubic inches. How many baseballs would

“fit” inside a basketball? a. 46 c. 27 b. 53 d. 35
Mathematics
2 answers:
Law Incorporation [45]3 years ago
8 0
<span>Divide the larger volume by the smaller volume. Use a calculator.</span>
ehidna [41]3 years ago
6 0
I hope this helps you




448.92/12.77



44892/1277



37.1
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5.65 times 3.4 estimate
lidiya [134]
We can round 5.65 up to 6 and 3.4 down to 3.
3 * 6 = 18
now let's see how close our estimate is to the real answer
5.65 * 3.4 = 19.21
Our answer was pretty close to the real answer!

Hope I helped!
~ Zoe
3 0
4 years ago
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John bought a used truck for $4,500. He made an agreement with the dealer to put $1,500 down and make payments of $350 for the n
kumpel [21]
I think it is D But it could also be A
7 0
3 years ago
The circumstance of a circle is 36pi inches what is the radius for this circle
frez [133]

Circumference = 2pi*r

r = radius

In our problem,

circumference = 36pi

r = ?

Let's plug our values into the formula above.

36pi = 2pi * r

Divide both sides by 2pi

18 = r

4 0
3 years ago
3. Let sin 21 = 0.358. What is the value of x for cos x=0.358?
r-ruslan [8.4K]

Answer:

\cos(69)

Step-by-step explanation:

We will use the co-function identity for sine, where:

\sin(x)=\cos(90-x)

Since sin(21)=0.358=cos(x). This means that:

21=90-x

Solve for x. Add x to both sides:

x+21=90

Subtract 21 from both sides:

x=69\textdegree

Hence:

\sin(21)=\cos(69)\approx0.358

6 0
3 years ago
Inverse laplace of [(1/s^2)-(48/s^5)]
Katen [24]
**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform F(s) and G(s), then \mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}

Given that \dfrac{1}{s^2} = \dfrac{1!}{s^2} and -\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}

From Table of Laplace Transform, you have \mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}} and hence \mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}.

Hope this helps...
7 0
3 years ago
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