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2 years ago

6. Which expression is equivalent to (6. p) + 3?

Mathematics

1 answer:

Gelneren [198K]2 years ago

5 0

Answer:

3 + (p ⋅ 6)

Step-by-step explanation:

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Which expression is equivalent to (2g^5)^3/(4h^2)^3

9966 [12]

$\cfrac{(2g^5)^3}{(4h^2)^3} = \cfrac{2^3 \cdot g^{5\cdot3}}{4^3 \cdot h^{2\cdot3}} = \cfrac{8 g^{15}}{64 h^{6}} =\cfrac{g^{15}}{8h^{6}}$

First option

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2 years ago

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A polygon is translated 3 units to the right and 4 units down, then it’s reflected across the x-axis, and then it’s dilated by a

Marrrta [24]

Answer:

$(x,y)\rightarrow (4x+12,-4y+16)$

Step-by-step explanation:

1 transformation - translation 3 units to the right and 4 units down. This translation has the rule

$(x,y)\rightarrow (x+3,y-4)$

2 transformation - reflection across the x-axis. This reflection has the rule

$(x,y)\rightarrow (x,-y)$

3 transformation - dilation by a factor of 4 with the origin as the center of dilation. This dilation has the rule

$(x,y)\rightarrow (4x,4y)$

Now, the sequence of these three transformation has the rule

$(x,y)\rightarrow\limits^{\text{1st transformation}} (x+3,y-4)\rightarrow\limits^{\text{2nd transformation}}(x+3,-(y-4))\rightarrow\limits^{\text{3rd transformation}} (4(x+3),4(-(y-4)))\\ \\(x,y)\rightarrow (4x+12,-4y+16)$

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2 years ago

Find the quotient 288/16 which correct A:13 R 20 B:14 R 4 C:103 R 20 D:104 R 4

Andru [333]

Answer:

the answer is C

Step-by-step explanation:

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2 years ago

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Joseph invests money in an account paying a simple interest of 2.9% per year. If he invests $60 and no money will be added or re

777dan777 [17]

Hi there! I’m here to help you out with your question.

It costs Joseph 2.9% per year in interest.

This is a one step process, do 60 (money he invests) x .029 (interest) and you should get $1.74.

Add $1.74 + $60, and your answer is $61.74.

Thank you and I hope this helped you.

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2 years ago

50℅ of what number is 7.5? Help please!!

Makovka662 [10]

Let the number be y.
so,
50% × y = 7.5
move 50% to the other side
y=7.5÷50%
therefore y=15

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2 years ago

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