Answer:
The volume of paint that covers the whole car is
gallons or one and a quarter gallons.
Step-by-step explanation:
Let the volume of paint required to cover whole car be 'x'.
Given:
Volume of paint to cover one-fifth of a car = one-fourth of a gallon.
We use unitary method to determine the volume of paint required.
∵ One-fifth of a car requires paint = ![\frac{1}{4}\ gal](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7D%5C%20gal)
∴ Whole volume of car requires paint = ![\frac{\frac{1}{4}}{\frac{1}{5}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cfrac%7B1%7D%7B4%7D%7D%7B%5Cfrac%7B1%7D%7B5%7D%7D)
![=\frac{1}{4}\times \frac{5}{1}\\=\frac{5}{4}\\=1\frac{1}{4}\ gal](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5Cfrac%7B5%7D%7B1%7D%5C%5C%3D%5Cfrac%7B5%7D%7B4%7D%5C%5C%3D1%5Cfrac%7B1%7D%7B4%7D%5C%20gal)
Therefore, the volume of paint that covers the whole car is
gallons or one and a quarter gallons.
Answer:
D) feet²
Step-by-step explanation:
Area = side × side
= feet × feet
= feet²
Let's assume we were given 2 numbers: 15 and 30. Their sum is:
![15+30 = 45](https://tex.z-dn.net/?f=%2015%2B30%20%3D%2045%20)
We want to express it as the product of GCF and another sum.
15 is divisible by: 1, 3, 5, 15
30 is divisible by: 1, 2, 3, 5, 6, 10, 15, 30
The greatest number that appears in 2 series is 15.
![\frac{15}{15} = 1](https://tex.z-dn.net/?f=%20%5Cfrac%7B15%7D%7B15%7D%20%3D%201%20)
![\frac{30}{15} = 2](https://tex.z-dn.net/?f=%20%5Cfrac%7B30%7D%7B15%7D%20%3D%202%20)
In this case sum of two numbers can always be written as:
Answer:
V_rocket = 44
ft^3 = 138.2 ft^3
Step-by-step explanation:
Volume of cone
V_cone = (1/3)*(
*(radius^2))*h
radius = 2ft
h = height of the cone = 3 ft
pi = 3.1416
V_cone = 4
ft^3
Volume cylinder
V_cylin = (
*(radius^2))*h_cyl
h_cyl = height of the cylinder = 10 ft
V_cylin = 40
ft^3
Volume rocket
V_rocket = V_cone + V_cylin = 4
ft^3 + 40
ft^3
V_rocket = 44
ft^3 = 138.2 ft^3
Answer: 20 feet by 20 feet.
Step-by-step explanation:
1. You know that
is the original measure of one side. Since it is a square, all its sides are equal.
2. Therefore, to solve the problem you only need to solve for
, as you can see below:
- You have that:
![(a-b)^{2}=a^{2}-2ab+b^{2}](https://tex.z-dn.net/?f=%28a-b%29%5E%7B2%7D%3Da%5E%7B2%7D-2ab%2Bb%5E%7B2%7D)
- Then, you obtain the following quadratic equation:
![(x-8)^{2} =144\\x^{2}-2(x)(8)+8^{2}=144\\x^{2}-16x-80=0](https://tex.z-dn.net/?f=%28x-8%29%5E%7B2%7D%20%3D144%5C%5Cx%5E%7B2%7D-2%28x%29%288%29%2B8%5E%7B2%7D%3D144%5C%5Cx%5E%7B2%7D-16x-80%3D0)
- Apply the Quadratic formula to solve it:
![x=\frac{-b+/-\sqrt{b^{2}-4ac}}{2a}\\a=1\\b=-16\\c=-80](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%2B%2F-%5Csqrt%7Bb%5E%7B2%7D-4ac%7D%7D%7B2a%7D%5C%5Ca%3D1%5C%5Cb%3D-16%5C%5Cc%3D-80)
- Then, you obtain:
![x_1=20\\x_2=-4](https://tex.z-dn.net/?f=x_1%3D20%5C%5Cx_2%3D-4)
3. The dimensions cannot be negative, therefore, the answer is: 20 feet by 20 feet.