As we can see in the shown figure:
1) I and J are parallel
2) I and g are perpendicular because we have constructed the steps to make the perpendicular to the line I
3) J and g are perpendiculars because I and J are parallel and I perpendiculars to g
Answer:
y = 1/2x + 3
O f(x)= 1/2x+3
Step-by-step explanation:
f(x) = 2x - 6
y = 2x - 6
x = 2y - 6
2y = x + 6
y = (x + 6)/2
y = 1/2x + 3
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>
Answer:
B. x = 2
Step-by-step explanation:
Answer:
x = 2
Step-by-step explanation:
check by substituting each given value of x into the function
x = 0 ⇒ y = × 0² = 0
hence (0, 0) is correct
x = 2 ⇒ y = × 4 = 3
hence (2, 4 ) is incorrect
x = 3 ⇒ y = × 9 = 6
hence (3, 6 ) is correct
x = 5 ⇒ y = × 25 = 18
hence (5, 18 ) is correct
