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3 years ago

Please help! Correct answers only please!

Mathematics

1 answer:

Maurinko [17]3 years ago

8 0

Answer:

B. is your answer

Step-by-step explanation:

V=whl=7·6.5·7=318.5

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5(m+4)=-2(-4-m)+3<br> help please

Degger [83]

Answer:

m = -3

Step-by-step explanation:

5m + 20 = 8 + 2m +3

5m + 20 = 11 +2m

3m + 20 = 11

3m = -9

m = -3

3 0

3 years ago

Can someone help me and my friend on this problem?

irina [24]

Absolute value was $55. I don't really get the problem...

4 0

3 years ago

How do you write 0.12 as a fraction

Arturiano [62]

[ Answer ]

$\boxed{\frac{3}{25} }$

[ Explanation ]

Rewrite 0.12 As A Fraction

-----------------------------

Rewrite Decimal As A Fraction With 1 As A Denominator

0.12 = $\frac{0.12}{1}$

Multiply To Remove Decimal Places

$\frac{0.12}{1}$ · $\frac{100}{100}$ = $\frac{12}{100}$

Find GCF, Reduce Fraction

$\frac{12 \ / \ 4}{100 \ / \ 4}$ = $\frac{3}{25}$

0.12 = $\frac{3}{25}$

$\boxed{\boxed{[ \ Eclipsed \ ]}}$

7 0

3 years ago

A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. An industrial tank of this shape must h

mestny [16]

Answer:

Radius =6.518 feet

Height = 26.074 feet

Step-by-step explanation:

The Volume of the Solid formed = Volume of the two Hemisphere + Volume of the Cylinder

Volume of a Hemisphere $=\frac{2}{3}\pi r^3$

Volume of a Cylinder $=\pi r^2 h$

Therefore:

The Volume of the Solid formed

$=2(\frac{2}{3}\pi r^3)+\pi r^2 h\\\frac{4}{3}\pi r^3+\pi r^2 h=4640\\\pi r^2(\frac{4r}{3}+ h)=4640\\\frac{4r}{3}+ h =\frac{4640}{\pi r^2} \\h=\frac{4640}{\pi r^2}-\frac{4r}{3}$

Area of the Hemisphere = $2\pi r^2$

Curved Surface Area of the Cylinder = $2\pi rh$

Total Surface Area=

$2\pi r^2+2\pi r^2+2\pi rh\\=4\pi r^2+2\pi rh$

Cost of the Hemispherical Ends = 2 X Cost of the surface area of the sides.

Therefore total Cost, C

$=2(4\pi r^2)+2\pi rh\\C=8\pi r^2+2\pi rh$

Recall: $h=\frac{4640}{\pi r^2}-\frac{4r}{3}$

Therefore:

$C=8\pi r^2+2\pi r(\frac{4640}{\pi r^2}-\frac{4r}{3})\\C=8\pi r^2+\frac{9280}{r}-\frac{8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{24\pi r^2-8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{16\pi r^2}{3}\\C=\frac{27840+16\pi r^3}{3r}$