Answer:
![\sqrt{145}](https://tex.z-dn.net/?f=%5Csqrt%7B145%7D)
Step-by-step explanation:
The distance formula states that the distance between two points
and
is
.
The two points we have are
and
. Plugging these numbers into the distance formula, we have
.
Simplifying with order of operations, first using the distributive property, gives
.
Squaring and adding gives
![\sqrt{(4+4)^2+(6+3)^2}=\sqrt{8^2+9^2}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\sqrt{64+81}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\boxed{\sqrt{145}},](https://tex.z-dn.net/?f=%5Csqrt%7B%284%2B4%29%5E2%2B%286%2B3%29%5E2%7D%3D%5Csqrt%7B8%5E2%2B9%5E2%7D%5C%5C~~~~~~~~~~~~~~~~~~~~~~~~~~~~%3D%5Csqrt%7B64%2B81%7D%5C%5C~~~~~~~~~~~~~~~~~~~~~~~~~~~~%3D%5Cboxed%7B%5Csqrt%7B145%7D%7D%2C)
which is the answer in simplest form. This also rounds to about 12.04.
Answer:
28 quarters and 12 nickles
Step-by-step explanation:
How many quarters goes into 7 dollars? 28, because there is 4 in each dollar.
How many nickles are in 60 cents? 12, because 5 times 12 is 60.
Answer:
The correct answer is B.
Step-by-step explanation:
In order to find this, calculate out the discriminant for each of the following equations. If the discriminant is a perfect square, then it can be factored.
Discriminant = b^2 - 4ac
The only of the equations that does not yield a perfect square is B. The work for it is done below for you.
Discriminant = b^2 - 4ac
Discriminant = 7^2 - 4(2)(-5)
Discriminant = 49 + 40
Discriminant = 89
Since 89 is not a perfect square, we cannot factor this.
Answer:
18
Step-by-step explanation:
10/0.5 = 20
20 * 0.9 = 18
<u>We are given</u>
- Radius of Earth; 6.4 x 100 meters = 640 meters
Clearly, the shape of the earth is a sphere. Thus, to determine the volume of the earth, we will use a formula that determines the volume of a sphere.
![\implies \text{Volume of sphere =} \ \dfrac{4\pi r^{3}}{3}](https://tex.z-dn.net/?f=%5Cimplies%20%5Ctext%7BVolume%20of%20sphere%20%3D%7D%20%5C%20%20%20%5Cdfrac%7B4%5Cpi%20r%5E%7B3%7D%7D%7B3%7D)
When we substitute the radius in the formula, we get;
![\implies\text{Volume of sphere} = \dfrac{4\pi (640)^{3}}{3}](https://tex.z-dn.net/?f=%5Cimplies%5Ctext%7BVolume%20of%20sphere%7D%20%3D%20%5Cdfrac%7B4%5Cpi%20%28640%29%5E%7B3%7D%7D%7B3%7D)
![\implies\text{Volume of sphere} = \dfrac{4\pi (640)(640)(640)}{3}](https://tex.z-dn.net/?f=%5Cimplies%5Ctext%7BVolume%20of%20sphere%7D%20%3D%20%5Cdfrac%7B4%5Cpi%20%28640%29%28640%29%28640%29%7D%7B3%7D)
Take π as 3.14
![\implies\text{Volume of sphere} = \dfrac{4\pi (640)(640)(640)}{3}](https://tex.z-dn.net/?f=%5Cimplies%5Ctext%7BVolume%20of%20sphere%7D%20%3D%20%5Cdfrac%7B4%5Cpi%20%28640%29%28640%29%28640%29%7D%7B3%7D)
![\implies \text{Volume of sphere} = \dfrac{4(3.14)(640)(640)(640)}{3}](https://tex.z-dn.net/?f=%5Cimplies%20%5Ctext%7BVolume%20of%20sphere%7D%20%3D%20%5Cdfrac%7B4%283.14%29%28640%29%28640%29%28640%29%7D%7B3%7D)
Simplify the numerator;
![\implies \text{Volume of sphere} = \dfrac{4(3.14)(640)(640)(640)}{3}](https://tex.z-dn.net/?f=%5Cimplies%20%5Ctext%7BVolume%20of%20sphere%7D%20%3D%20%5Cdfrac%7B4%283.14%29%28640%29%28640%29%28640%29%7D%7B3%7D)
![\implies \text{Volume of sphere} = \dfrac{3292528640}{3}](https://tex.z-dn.net/?f=%5Cimplies%20%5Ctext%7BVolume%20of%20sphere%7D%20%3D%20%5Cdfrac%7B3292528640%7D%7B3%7D)
Divide the numerator by 3;
![\implies \text{Volume of sphere} = \dfrac{3292528640}{3}](https://tex.z-dn.net/?f=%5Cimplies%20%5Ctext%7BVolume%20of%20sphere%7D%20%3D%20%5Cdfrac%7B3292528640%7D%7B3%7D)
![\implies \text{Volume of sphere} = \boxed{1097509546.67 \ \text{m}^{3} } \ \ \ (\text{Estimated})](https://tex.z-dn.net/?f=%5Cimplies%20%5Ctext%7BVolume%20of%20sphere%7D%20%3D%20%5Cboxed%7B1097509546.67%20%5C%20%5Ctext%7Bm%7D%5E%7B3%7D%20%7D%20%5C%20%5C%20%5C%20%28%5Ctext%7BEstimated%7D%29)