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3 years ago

Seventeen minus the quotient of twelve and six is the equivalent to which of the following?

Mathematics

1 answer:

KATRIN_1 [288]3 years ago

3 0

Answer:

It should be the first on the list 17 - 12/6

Step-by-step explanation:

if I replace the words in your sentence, you have

17 minus the 12/6

get

17 - 12/6

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The image of the point (-6, -2)under a translation is (−7,0). Find the coordinates of the image of the point (7,0) under the sam

lesya692 [45]

Answer:

hello : solutio ( 6 ; 2)

Step-by-step explanation:

the translation is : x' = x + a ....(*)

y' = y + b ...(**)

calculate the coordinates ( a ; b) of vectro by the translation

you have : x' = -7 y' = 0 x = -6 y = -2

Substitute in (*) , (***) : -7 = -6 +a

0 = - 2 + b

a = -1 and b= 2

so : x' = x - 1

y' = y + 2

calculate now the coordinates of the image of the point (7,0) under the same translation.

x = 7 and b = 0

so : x' = 7-1 = 6 and y' = 0 + 2 = 2

the image is : ( 6 ; 2 )

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3 years ago

A cone has a volume of 12 cubic inches. What is the volume of a cylinder that the cone fits exactly inside of?

Sunny_sXe [5.5K]

11 cubic inches yahhhh

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3 years ago

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Andrew lives at an altitude of 135 feet, and his friend Lisa lives at an altitude of -10 feet. blank residence is higher than bl

tatiyna

Answer:

Andrews residence is higher than Lisa's residence

Step-by-step explanation:

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3 years ago

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How to divide 83.7 by 0.03

Mandarinka [93]

The answer would be 2790

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3 years ago

Which are the solutions of the quadratic equation? x2 = –5x – 3 –5, 0 StartFraction negative 5 minus StartRoot 13 EndRoot Over 2

boyakko [2]

<u>Given</u>:

The quadratic equation is $x^{2}=-5 x-3$

We need to determine the solutions of the quadratic equation.

<u>Solution</u>:

Let us solve the equation to determine the value of x.

Adding both sides of the equation by 5x and 3, we get;

$x^{2}+5 x+3=0$

The solution of the equation can be determined using quadratic formula.

Thus, we get;

$x=\frac{-5 \pm \sqrt{5^{2}-4 \cdot 1 \cdot 3}}{2 \cdot 1}$

$x=\frac{-5 \pm \sqrt{25-12}}{2 }$

$x=\frac{-5 \pm \sqrt{13}}{2 }$

Thus, the two roots of the equation are $x=\frac{-5 + \sqrt{13}}{2 }$ and $x=\frac{-5- \sqrt{13}}{2 }$

Hence, the solutions of the equation are $x=\frac{-5 + \sqrt{13}}{2 }$ and $x=\frac{-5- \sqrt{13}}{2 }$

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3 years ago

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