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andriy [413]
4 years ago
13

114 miles on 4.5 gallons

Mathematics
1 answer:
ivann1987 [24]4 years ago
5 0

Answer:

Miles using 4.5 gallons = 114 miles

Miles using 1 gallon

(114 miles / 4.5 gallons)

32 miles / gallon

Step-by-step explanation:

sorry if wrong

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I just wanted to see if this was 8 or negative 8 :)​
nirvana33 [79]

Answer:

It's a positive 8.

Step-by-step explanation:

You can't have a negative amount of money.

4 0
3 years ago
If AB=48 and BC=22, then AC=?
alexandr1967 [171]

If A, B and C are collinear, then

1) if B is between A and C:

AC = AB + BC

AC = 48 + 22 = 70

2) if C is between A and B:

AB = AC + BC

48 = AC + 22 |-22

AC = 26

3) if A is between B and C:

BC = AB + AC

22 = 48 + AC |-48

AC = - 26 < 0 FALSE

Answer:

if B is between A and C, then AC = 70

if C is between A and B, then AC = 26

4 0
3 years ago
Read 2 more answers
How to factor 1-49c^2
Dima020 [189]
Ok so this is a difference of 2 perfect squares
1=1^2
49c^2=(7c)^2
to factor you remember
a^2-b^2=(a+b)(a-b) so

1^2-(7c)^2=(1+7c)(1-7c)
5 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}

The minimum cost occurs at the point where the derivative equals zero.

C^{'}=\frac{-27840+32\pi r^3}{3r^2}

When \:C^{'}=0

-27840+32\pi r^3=0\\27840=32\pi r^3\\r^3=27840 \div 32\pi=276.9296\\r=\sqrt[3]{276.9296} =6.518

Recall:

h=\frac{4640}{\pi r^2}-\frac{4r}{3}\\h=\frac{4640}{\pi*6.518^2}-\frac{4*6.518}{3}\\h=26.074 feet

Therefore, the dimensions that will minimize the cost are:

Radius =6.518 feet

Height = 26.074 feet

5 0
3 years ago
The Great Wall is more than 20,000 kilometers long about 3/4 of it is considered properly preserved
yaroslaw [1]
97. The great wall is more than 20 000 kilometers long. And about ¾ of it is considered as properly preserved. I guess you are asking about the value of ¾ which is properly preserved in the said length of the great wall. => ¾ = 0.75 Thus, the formula would be like this: => 20 000 * .75 = 15 000 Thus, the ¾ value of great wall that is preserved is equals to 15 000 kilometers long,
7 0
3 years ago
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