(x + 3)^2 + (x + 4)^2
= x^2 + 6x + 9 + x^2 + 8x + 16
= 2x^2 + 14x + 25
= 2(x^2 + 7x) + 25
= 2[(x + 7/2)^2 - 49/4] + 25
= 2(x + 7/2)^2 - 98/4 + 25
= 2(x + 7/2)^2 + 1/2
Its B
Answer:
the x intercept is (0,100) and the y intercept is (0,0)
Step-by-step explanation:
g(x)=1/20x(x-100)
g(0)=1/20(0)(0-100)
g(0)=1/20*0(0-100)
g(0)=100
Answer:
![\sqrt[3]{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B3%7D)
Step-by-step explanation:
Our expression is: ![\frac{1}{3} \sqrt[3]{81}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%20%5Csqrt%5B3%5D%7B81%7D)
.
Let's focus on the cube root of 81 first. What's the prime factorisation of 81? It's simply: 3 * 3 * 3 * 3, or 
. Put this in for 81:
![\sqrt[3]{81} =\sqrt[3]{3^3*3}=\sqrt[3]{3^3} *\sqrt[3]{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B81%7D%20%3D%5Csqrt%5B3%5D%7B3%5E3%2A3%7D%3D%5Csqrt%5B3%5D%7B3%5E3%7D%20%2A%5Csqrt%5B3%5D%7B3%7D)
We know that the cube root of 3 cubed will cancel out to become 3, but the cube root of 3 cannot be further simplified, so we keep that. Our outcome is then:
![\sqrt[3]{3^3} *\sqrt[3]{3}=3\sqrt[3]{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B3%5E3%7D%20%2A%5Csqrt%5B3%5D%7B3%7D%3D3%5Csqrt%5B3%5D%7B3%7D)
Now, let's multiply this by 1/3, as shown in the original problem:
![\frac{1}{3}* 3\sqrt[3]{3}=\sqrt[3]{3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%2A%203%5Csqrt%5B3%5D%7B3%7D%3D%5Csqrt%5B3%5D%7B3%7D)
Thus, the answer is .
<em>~ an aesthetics lover</em>
The answer is B) multiply 3 by 3. With that 2 on top, that doesn't mean 3 times 2, that means multiply 3 by itself, however many times the number says, if that makes sense. So if there's a 2 above the 3, then that would be 3*3. If it was a 5 instead of a two, that would mean 3*3*3*3*3. Not 3*5.
I think it’s always because the absolute value is always positive.
