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3 years ago

Six more than a quotient of a number and three is greater than 14

Mathematics

2 answers:

Mashcka [7]3 years ago

6 0

Its actually really simple. x/3 + 6> 14 , or rather, x divided by 3 + 6 is greater than 14

Lorico [155]3 years ago

3 0

Remark

You sure do have to read that a couple of times.

The quotient is an answer after a division takes place So you are dividing x/3

When you have figured out that phrase you add six onto it.

(x/3) + 6 > 14 Subtract 6 from both sides.

(x/3) +6 - 6 > 14 - 6

(x/3) > 8 Now multiply by 3

x > 8*3

x > 24 <<<<<< Answer

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Write an equation of the perpendicular bisector of the line segment whose endpoints are (−1,1) and (7,−5)

icang [17]

The equation is $y = \frac{3}{2} x - \frac{11}{2}$

<u>Explanation:</u>

We have to first find the mid-point of the segment, the formula for which is

$(\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2} )$

So, the midpoint will be $(\frac{-1+7}{2} , \frac{1-5}{2} )\\\\$

= $(3,-2)$

It is the point at which the segment will be bisected.

Since we are finding a perpendicular bisector, we must determine what slope is perpendicular to that of the existing segment. To determine the segment's slope, we use the slope formula $\frac{y_2-y_1}{x_2-x_1}$

The slope is $\frac{-5-1}{7+1}$ = $-\frac{2}{3}$

Perpendicular lines have opposite and reciprocal slopes. The opposite reciprocal of $-\frac{2}{3}$ is $\frac{3}{2}$

To write an equation, substitute the values in y = mx + c

WHere,

y = -1

x = 3

m = 3/2

Solving for c:

$-1 = \frac{3}{2} X 3 + c\\\\-1 = \frac{9}{2}+c\\ \\c = \frac{-2-9}{2} \\\\c = \frac{-11}{2}$

Thus, the equation becomes:

$y = \frac{3}{2} x - \frac{11}{2}$

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Step-by-step explanation:

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