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zzz [600]
3 years ago
9

The volume of this math problem​

Mathematics
2 answers:
Brrunno [24]3 years ago
5 0
Your answer will be “1200”
DerKrebs [107]3 years ago
3 0

The volume of the math problem is 1200 yd.

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Find the variable pls help
Semenov [28]

Answers:

1. a=600

2. b=8.5

3. c=1

4. d= 5/3

5. e= *error*

6. f=20

7. g=0.01

8= h=4/7

8 0
3 years ago
Read 2 more answers
If g(x) = 4(x - 2)2 - 4, complete the following statements.
Digiron [165]

Answer:

x = 10

Step-by-step explanation:

Please, use " ^ " to denote exponentiation:    g(x) = 4(x - 2)^2 - 4

The vertex is located at (2, -4) which numbers come directly from the '2' in (x - 2) and the "-4" at the end of the equation.

Where is the function f(x) that you mentioned?

The axis of symmetry here is the vertical line that passes through the vertex.  Its equation is x = 10.

7 0
4 years ago
Read 2 more answers
If the ratio of radius of two spheres is 4:7, the ratio of their volume is?​
Kazeer [188]

Answer:

64 : 343

Step-by-step explanation:

First use the radii to find the volume

1) Radius of first sphere is 4 (taken from 4:7 ratio)

   Insert it into the equation for volume of a sphere: V=4 /3πr^3

   V = (4/3)(π)(4^3)

   V = (4/3)(π)(64)

   V = 256/3 π

   Volume of the first sphere = 256/3 π

2) Radius of the second sphere is 7 (also taken from 4:7 ratio)

   Insert it into the equation for volume of a sphere: V=4 /3πr^3

   V = (4/3)(π)(7^3)

   V = (4/3)(π)(343)

   V = 1372/3 π

   Volume of the second sphere = 1372/3 π

Next, calculate the ratio by dividing the two numbers

256/3 π ÷ 1372/3 π

Answer should be 64 : 343

The simple way to do this problem is to just cube the numbers:  

4:7 becomes 4^3 : 7^3 = 64 : 343

Either way works.

6 0
3 years ago
What are three ways to name a line with 5 points?
sertanlavr [38]
EJ
EGJ
EFFHJ
Hope I helped
4 0
4 years ago
If
likoan [24]

It's easy to show that 7\tan(4x) is strictly increasing on x\in\left[0,\frac\pi8\right]. This means

M = \max \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/12} = 7\sqrt3

and

m = \min \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/16} = 7

Then the integral is bounded by

\displaystyle 7\left(\frac\pi{12} - \frac\pi{16}\right) \le \int_{\pi/16}^{\pi/12} 7\tan(4x) \, dx \le 7\sqrt3 \left(\frac\pi{12} - \frac\pi{16}\right)

\implies \displaystyle \boxed{\frac{7\pi}{48}} \le \int_{\pi/16}^{\pi/12} 7\tan(4x) \, dx \le \boxed{\frac{7\sqrt3\,\pi}{48}}

7 0
2 years ago
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