Picture graphs usually show amounts using...pictures! So if you looked
at the symbols, you can likely guess the amounts. It might show how many
students preferred hamburgers by using a hamburger to represent every 5
students. Just by quickly glancing at a picture graph, you can usually
tell which result is the greatest.
Answer:
5.4375 Pounds (lb)
Step-by-step explanation:
By using the calculator, and dont call your self stupid we all need help sometimes
6/7, 12/14, 18,21, 24/28, 30/35, 36/42 in so on
Answer:
![4x^{3} y^{2} (\sqrt[3]{4 x y})](https://tex.z-dn.net/?f=4x%5E%7B3%7D%20y%5E%7B2%7D%20%28%5Csqrt%5B3%5D%7B4%20x%20y%7D%29)
Step-by-step explanation:
Another complex expression, let's simplify it step by step...
We'll start by re-writing 256 as 4^4
![\sqrt[3]{256 x^{10} y^{7} } = \sqrt[3]{4^{4} x^{10} y^{7} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B256%20x%5E%7B10%7D%20y%5E%7B7%7D%20%7D%20%3D%20%5Csqrt%5B3%5D%7B4%5E%7B4%7D%20x%5E%7B10%7D%20y%5E%7B7%7D%20%7D)
Then we'll extract the 4 from the cubic root. We will then subtract 3 from the exponent (4) to get to a simple 4 inside, and a 4 outside.
![\sqrt[3]{4^{4} x^{10} y^{7} } = 4 \sqrt[3]{4 x^{10} y^{7} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B4%5E%7B4%7D%20x%5E%7B10%7D%20y%5E%7B7%7D%20%7D%20%3D%204%20%5Csqrt%5B3%5D%7B4%20x%5E%7B10%7D%20y%5E%7B7%7D%20%7D)
Now, we have x^10, so if we divide the exponent by the root factor, we get 10/3 = 3 1/3, which means we will extract x^9 that will become x^3 outside and x will remain inside.
![4 \sqrt[3]{4 x^{10} y^{7} } = 4x^{3} \sqrt[3]{4 x y^{7} }](https://tex.z-dn.net/?f=4%20%5Csqrt%5B3%5D%7B4%20x%5E%7B10%7D%20y%5E%7B7%7D%20%7D%20%3D%204x%5E%7B3%7D%20%5Csqrt%5B3%5D%7B4%20x%20y%5E%7B7%7D%20%7D)
For the y's we have y^7 inside the cubic root, that means the true exponent is y^(7/3)... so we can extract y^2 and 1 y will remain inside.
![4x^{3} \sqrt[3]{4 x y^{7} } = 4x^{3} y^{2} \sqrt[3]{4 x y}](https://tex.z-dn.net/?f=4x%5E%7B3%7D%20%5Csqrt%5B3%5D%7B4%20x%20y%5E%7B7%7D%20%7D%20%3D%204x%5E%7B3%7D%20y%5E%7B2%7D%20%5Csqrt%5B3%5D%7B4%20x%20y%7D)
The answer is then:
![4x^{3} y^{2} \sqrt[3]{4 x y} = 4x^{3} y^{2} (\sqrt[3]{4 x y})](https://tex.z-dn.net/?f=4x%5E%7B3%7D%20y%5E%7B2%7D%20%5Csqrt%5B3%5D%7B4%20x%20y%7D%20%3D%204x%5E%7B3%7D%20y%5E%7B2%7D%20%28%5Csqrt%5B3%5D%7B4%20x%20y%7D%29)
If the endpoints of a diameter are (6,3) and (2,1) the midpoint is the center of the circle so:
(x,y)=((6+2)/2, (3+1)/2)=(4,2)
Now we need to find the radius....the diameter is:
d^2=(6-2)^2+(3-1)^2
d^2=16+4
d^2=20 since d=2r, r=d/2, and r^2=d^2/4 so
r^2=5
The standard form of the circle is (x-h)^2+(y-k)^2=r^2 and we know:
(h,k)=(4,2) from earlier so:
(x-4)^2+(y-2)^2=5
