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vichka [17]
2 years ago
11

Q={1,3,6,9,10,15} then rewrite in set builder method

Mathematics
1 answer:
AfilCa [17]2 years ago
4 0

Answer:

{x:x is a multiple of 3, x ≤ 15}

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(-7x+4)(7x+4)
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4 years ago
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Subtract the sum of (3/5+1/5 i) and (4/5-2/5 i) from (9/5-1/5 i)
lyudmila [28]

<em>The answer is -2/5.</em>

(3/5+1/5)+(4/5-2/5)-(9/5-1/5)=

(4/5+2/5)-8/5=

6/5-8/5=

-2/5.

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3 years ago
I’ll give brainliest!!! question in the picture
Bogdan [553]

Answer:

5a + 2b is the total amount Hattie spent

Step-by-step explanation:

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3 years ago
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Determine if the two lines are perpendicular. Explain how you know by calculating the slopes and comparing the slopes in 2-3 sen
statuscvo [17]

Answer:

The two lines are not perpendicular because the product of their slopes is not equal to -1

Step-by-step explanation:

The product of the slopes of the perpendicular line is -1

  • That means one of them is and additive and multiplicative inverse of the other
  • If the slope of one of them is m, then reciprocal m and change its sign, then the slope of the perpendicular is -\frac{1}{m}
  • The formula of the slope is m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

Let us find from the graph two points lie on each line and calculate the slopes of them and then find its product if the product is -1, then the two lines are perpendicular

From the graph

∵ The red line passes through points (4 , 0) and (0 , 8)

∴ x_{1} = 4 and x_{2} = 0

∴ y_{1} = 0 and y_{2} = 8

∵ m=\frac{8-0}{0-4}=\frac{8}{-4}=-2

∴ The slope of the red line is -2

∵ The blue line passes through points (5 , 5) and (0 , -5)

∴ x_{1} = 5 and x_{2} = 0

∴ y_{1} = 5 and y_{2} = -5

∵ m=\frac{-5-5}{0-5}=\frac{-10}{-5}=2

∴ The slope of the blue line is 2

∵ The products of the slopes of the two lines = -2 × 2 = -4

∴ The product of the slopes of the lines not equal -1

∴ The two lines are not perpendicular

6 0
3 years ago
The table shows the results of a survey of 400 random people on whether they like liquid soap, bar soap, or both. A 4-column tab
In-s [12.5K]

Answer: 100/400

<u>Step-by-step explanation:</u>

\begin{array}{l|c|c|c}&Liquid&Not\ Liquid&Total\\ Bar&200&100&300\\Not\ Bar&80&20&100\\Total&280&120&400\end{array}

\dfrac{Not\ Bar:Total}{Total:Total}\quad =\dfrac{100}{400}

4 0
3 years ago
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