Simplify,
$$(4^y+4^y+4^y+4^y)\cdot(3^y+3^y+3^y)$$
$$\left(A\right)\ 4^{4y}\cdot3^{3y}$$
$$(B)12^{y+1}$$
$$(C)16^y+9^y$$
$$(D)12^y$$
$$(E)4^y*12^y$$
The OA is B.
Can any explain this PS question Please? I don't understand it. Thanks.
Simplify $$(4^y+4^y+4^y+4^y)\cdot(3^y+3^y+3^y)$$
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4^y + 4^y + 4^y + 4^y = 4 * 4^y = 4^(y+1)LUANDATO wrote:Simplify,
$$(4^y+4^y+4^y+4^y)\cdot(3^y+3^y+3^y)$$
$$\left(A\right)\ 4^{4y}\cdot3^{3y}$$
$$(B)12^{y+1}$$
$$(C)16^y+9^y$$
$$(D)12^y$$
$$(E)4^y*12^y$$
The OA is B.
Can any explain this PS question Please? I don't understand it. Thanks.
3^y + 3^y + 3^y = 3 * 3^y = 3^(y + 1)
4^(y+1) * 3^(y+1) = 12^(y+1)
The answer is B
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Let y=-1.LUANDATO wrote:Simplify,
$$(4^y+4^y+4^y+4^y)\cdot(3^y+3^y+3^y)$$
$$\left(A\right)\ 4^{4y}\cdot3^{3y}$$
$$(B)12^{y+1}$$
$$(C)16^y+9^y$$
$$(D)12^y$$
$$(E)4^y*12^y$$
Plugging y=-1 into the given expression, we get:
(4¯¹ + 4¯¹ + 4¯¹ + 4¯¹)(3¯¹ + 3¯¹ + 3¯¹) = (1/4 + 1/4 + 1/4 + 1/4)(1/3 + 1/3 + 1/3) = (1)(1) = 1.
The target value is 1.
Now plug y=-1 into the answers to see which yields the target value of 1.
A: 4¯� + 3¯³ = 1/(4�) + 1/(3³) ≠1.
B: 12� = 1.
C: 16¯¹ + 9¯¹ = (1/16) + (1/9) ≠1.
D: 12¯¹ = 1/12 ≠1.
E: 4¯¹ * 12¯¹ = (1/4)(1/12) ≠1.
Only B works.
The correct answer is B.
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Hi LUANDATO,Simplify,
$$(4^y+4^y+4^y+4^y)\cdot(3^y+3^y+3^y)$$
$$\left(A\right)\ 4^{4y}\cdot3^{3y}$$
$$(B)12^{y+1}$$
$$(C)16^y+9^y$$
$$(D)12^y$$
$$(E)4^y*12^y$$
Let's take a look at your question.
$$(4^y+4^y+4^y+4^y)\cdot(3^y+3^y+3^y)$$
Factor out 4^y and 3^y
$$=\ 4^y\left(1+1+1+1\right).3^y\left(1+1+1\right)$$
$$=\ 4^y\left(4\right).3^y\left(3\right)$$
$$=\ 4^{y+1}.3^{y+1}$$
$$=\ \left(4\times3\right)^{y+1}$$
$$=\ 12^{y+1}$$
Therefore, Option B is correct.
Hope this helps.
I am available if you'd like any follow up.
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Note that 4^y is common to the first factor, and 3^y is common to the second factor. Thus, we have:LUANDATO wrote:Simplify,
$$(4^y+4^y+4^y+4^y)\cdot(3^y+3^y+3^y)$$
$$\left(A\right)\ 4^{4y}\cdot3^{3y}$$
$$(B)12^{y+1}$$
$$(C)16^y+9^y$$
$$(D)12^y$$
$$(E)4^y*12^y$$
(4^y + 4^y + 4^y + 4^y) x (3^y + 3^y + 3^y)
4^y(1 + 1 + 1 + 1) x 3^y(1 + 1 + 1)
4^y(4) x 3^y(3)
4^(y+1) x 3^(y+1)
12^(y+1)
Answer: B
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