Answer:
I don't exactly know what the question is here
Answer:
x = 30 and y = -34
Step-by-step explanation:
Given the following functions
(1/4)^(x+y) = 256... 1
log₄(x-y) = 3.... 2
From equation 2;
x-y = 4³
x-y = 64
x = 64 + y ... 3
Substitutw 3 into 1
From 1:
(1/4)^(x+y) = 256
(1/4)^(64+y+y) = 256
(1/4)^(64+2y) = 256
Take log₄ of both sides
64+2y log₄ (1/4) = log₄256
-(64+2y) = 4log₄4
-(64+2y) = 4
64+2y = -4
2y = -4 - 64
2y = -68
y = -34
Since
x = 64 + y .
x = 64 - 34
x = 30
Hence x = 30 and y = -34
Answer:
x<3 AND x>2
Step-by-step explanation:
First, 4x-4<8. Adding 4 to both sides, 4x<12, and dividing by 4, x<3.
Next, 9x+5>23. Subtracting 5 from both sides, 9x>18, and dividing by 9, x>2.
Therefore, x<3 AND x>2.
Answer:
<AFB
Step-by-step explanation:
Supplementary angles are the angles that's sum up to 180° which is called a straight angle.
In this question we see AFB is supplementary to DFA for DFB is the straight angle.
Answer:
![\sqrt[3]{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B3%7D)
Step-by-step explanation:
Our expression is:
.
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>