Answer:
-3a-4b+5
Step-by-step explanation:
(3a-6b+12)-(6a-2b+7)
3a-6b+12-6a+2b-7
3a-6a-6b+2b+12-7
-3a-4b+5
The answer is A, Hope this helps <3... can i have brainlyest?
Answer:
Final answer is ![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3Dx)
.
Step-by-step explanation:
Given problem is ![\sqrt[3]{x}\cdot\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D)
.
Now we need to simplify this problem.
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D)
Apply formula
![\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5Ep%7D%5Ccdot%5Csqrt%5Bn%5D%7Bx%5Eq%7D%3D%5Csqrt%5Bn%5D%7Bx%5E%7Bp%2Bq%7D%7D)
so we get:
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3D%5Csqrt%5B3%5D%7Bx%5E%7B1%2B2%7D%7D)
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3D%5Csqrt%5B3%5D%7Bx%5E%7B3%7D%7D)
Hence final answer is .
Answer:
Table 1 and 2 represent a function
Step-by-step explanation:
Given
<em>Table 1</em>
x 5 10 11
y 3 9 15
<em></em>
<em>Table 2</em>
x 5 10 11
y 3 9 9
<em>Table 3</em>
x 5 10 10
y 3 9 15
Required
Determine which of the tables represent that y is a function of x
For a relation to be a function; the x values must be unique.
In other words, each x value must not be repeated;
Having said that;
Analyzing Table 1
<em>Table 1</em>
x 5 10 11
y 3 9 15
<em></em>
Note that the x rows are unique as no value were repeated;
Hence, Table 1 is a function
<em>Table 2</em>
x 5 10 11
y 3 9 9
Note that the x rows are unique as no value were repeated;
Hence, Table 2 is a function
<em>Table 3</em>
x 5 10 10
y 3 9 15
Note that the x rows are not unique because 10 was repeated twice;
Hence, Table 3 is not a function
