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3 years ago

Plzzzz Answer As SOON AS POSSIBLE. What is the volume of the figure? Down below? it will be worth your while ;)

Mathematics

2 answers:

Greeley [361]3 years ago

6 0

The answer is 60 . 10+18+4 = 32
10+9+9 = 28 and your answer is 60

Makovka662 [10]3 years ago

4 0

Answer:

=60 cm

Step-by-step explanation:

i took the test

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Can someone please answer this question I’ll LITERALLY do ANNNYYTHHINNNGG Number 25

Paha777 [63]

Answer:

Step-by-step explanation:

15+ 4 1/2 + 6 3/4 = 26 1/4 in. and that is just 3 3/4 in. less 2 1/2 feet/ 30 in.

4 0

3 years ago

6. If you can run 1 mile in 8 minutes, how long will it take to run 4 miles (assuming you can maintain

Flauer [41]

Answer:

32 minutes

Step-by-step explanation:

all you have to do is know that the unit rate is 4 so multiply 8 and 4 to get 32

8 0

2 years ago

Read 2 more answers

What is 3√125x^12 A) 5x^2 B) 5x^4 C) 25x^2 D) 25x^4

Schach [20]

$\sqrt[3]{125x^{12}}=\sqrt[3]{125}\cdot\sqrt[3]{x^{4\cdot3}}=5\sqrt[3]{(x^4)^3}=5x^4\\\\\text{Answer:}\ B)\ 5x^4\\\\\text{Used:}\\\sqrt[n]{a\cdot b}=\sqrt[n]{a}\cdot\sqrt[n]{b}\\\\(a^n)^m=a^{n\cdot m}\\\\\sqrt[n]{a^n}=a$

5 0

3 years ago

Read 2 more answers

Find the 14th term of the geometric sequence 7, 28, 112,

Fed [463]

Answer:

a(14)=469762048

Step-by-step explanation:

geometric sequence= a(n)=a*r^(n-1)

we have:

a=7

r=28/7=4

So:

a(14)=7*4^(14-1)

=7*4^(13)

=469762048

hope this helps

6 0

3 years ago

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)= 6x^1/3 + 3x^4/3. You must justify

irina [24]

Answer:

x-coordinates of relative extrema = $\frac{-1}{2}$

x-coordinates of the inflexion points are 0, 1

Step-by-step explanation:

$f(x)=6x^{\frac{1}{3}}+3x^{\frac{4}{3}}$

Differentiate with respect to x

$f'(x)=6\left ( \frac{1}{3} \right )x^{\frac{-2}{3}}+3\left ( \frac{4}{3} \right )x^{\frac{1}{3}}=\frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}$

$f'(x)=0\Rightarrow \frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}=0\Rightarrow x=\frac{-1}{2}$

Differentiate f'(x) with respect to x

$f''(x)=2\left ( \frac{-2}{3} \right )x^{\frac{-5}{3}}+\frac{4}{3}x^{\frac{-2}{3}}=\frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}\\f''(x)=0\Rightarrow \frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}=0\Rightarrow x=1$

At x = $\frac{-1}{2}$ ,

$f''\left ( \frac{-1}{2} \right )=\frac{4\left ( -1+4\left ( \frac{-1}{2} \right ) \right )}{3\left ( \frac{-1}{2} \right )^{\frac{5}{3}}}>0$

We know that if $f''(a)>0$ then x = a is a point of minima.

So, $x=\frac{-1}{2}$ is a point of minima.

For inflexion points:

Inflexion points are the points at which f''(x) = 0 or f''(x) is not defined.

So, x-coordinates of the inflexion points are 0, 1

7 0

3 years ago