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3 years ago

What operation do we use to "scale up" an

Mathematics

1 answer:

ycow [4]3 years ago

6 0

I think it's multiplication

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What is the difference of (6d+2)-(3d -3)

lapo4ka [179]

Answer:

3d+5

Step-by-step explanation:

Say x is the difference

Solve for x.

6d+2−(3d−3)=x

Step 1: Flip the equation.

x=6d+2−(3d−3)

Step 2: Simplify

x=(6d-3d)+(2+3)

x=3d+5

7 0

3 years ago

Factor<br><img src="https://tex.z-dn.net/?f=16%20%7Bs%7D%5E%7B2%7D%20%20%2B%2040s%20%2B%2024" id="TexFormula1" title="16 {s}^{2}

Salsk061 [2.6K]

Hi there!

To factor this polynomial, we would have to divide the whole equation by 8.
8 is the Greatest Common Factor of 16, 40, and 24.

8(2s² + 5s + 3)
now factor it again

8(2s + 3)(s + 1)

So your answer is 8(2s + 3)(s + 1).
Hope this helped!

7 0

3 years ago

2y to the power 4 x 5y to the power 3

iogann1982 [59]

Answer:

(2*5y^7)

Step-by-step explanation:

STEP

Equation at the end of step 1

(2y4 • 5) • y3

STEP

Equation at the end of step 2

(2•5y4) • y3

STEP

Multiplying exponential expressions :

3.1 y4 multiplied by y3 = y(4 + 3) = y7

Final result :

(2•5y7)

5 0

3 years ago

Determine whether CS and KP are parallel, perpendicular, or neither. Show work.

dezoksy [38]

We need to compare the slopes....

CS
(1,12)(5,4)
slope = (4 - 12) / (5 - 1) = -8/4 = -2...so the slope of CS is - 2

KP
(1,9)(6,-6)
slope = (-6 - 9) / (6 - 1) = - 15/5 = -3...so the slope is -3

if the slopes would have been the same, it would be parallel because parallel lines have the same slope. if the slopes would have had negative reciprocal slopes, then it would be perpendicular. But since the slopes are neither of those...ur answer is neither.

6 0

4 years ago

Which expressions are equivalent to 42-5? Choose all the correct answers. Show your work in arriving at your answers.

frozen [14]

Answer:

B, C, E

Step-by-step explanation:

<u />

<u>Exponent Rules</u>

$a^b \cdot a^c=a^{b+c}$

$\dfrac{a^b}{a^c}=a^{b-c}$

$a^n \cdot b^n=(ab)^n$

$a^0=1$

$\dfrac{1}{a^b}=a^{-b}$

$(a^b)^c=a^{bc}$

$\textsf{A}\quad42^{-3} \cdot 42^8=42^{-3+8}=42^5$

$\textsf{B}\quad \dfrac{42^{-7}}{42^{-2}}=42^{-7-(-2)}=42^{-5}$

$\textsf{C}\quad \dfrac{6^{-5}}{7^5}=6^{-5} \cdot 7^{-5}=(6 \cdot 7)^{-5}=42^{-5}$

$\textsf{D}\quad 42^0 \cdot 42^5=1 \cdot 42^5=42^5$

$\textsf{E}\quad \left(\dfrac{1}{42^{-5}}\right)^{-1}=(42^5)^{-1}=42^{-5}$

$\textsf{F}\quad \left(\dfrac{6^{-5}}{7^5}\right)^{-1}=\left(6^{-5} \cdot 7^{-5}\right)^{-1}=\left(\left(6 \cdot 7\right)^{-5}\right)^{-1}=42^{-5(-1)}=42^5$

3 0

2 years ago

Read 2 more answers