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3 years ago

Which of the following is in the solution set of the inequality –28 < v – 16

2 answers:

8 0

You haven't mentioned what 'the following' are...

Anyway:
-28 < v - 16

Add 16 to both sides of the inequality:
-12 < v

So we know that v must be a number larger than -12

Hope it helped :)

exis [7]3 years ago

6 0

Let's solve your inequality step-by-step.−28<v−16
Step 1: Flip the equation.v−16>−28
Step 2: Add 16 to both sides.v−16+16>−28+16
v>−12
Answer:v>−12

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Eight Divided by 534

valentina_108 [34]

Answer : 0.0149812734

8 ÷ 534 = 0.0149812734
Hope this Helps :)

3 0

3 years ago

Please answer I am begging !!!

dlinn [17]

Answer:

z=8,154

Step-by-step explanation:

To find the answer for this is you take 21,137 and subtract 12,983 from it.

21,137-12,983=8,154

12,983 + z = 21,137

12,983 + 8,154 = 21,137

3 0

3 years ago

Which Answer is closet to the difference?

bearhunter [10]

Well the difference is -43 so I'm pretty sure it's gonna be A

5 0

3 years ago

Read 2 more answers

∆ABC has vertices A(–2, 0), B(0, 8), and C(4, 2)

Natali [406]

Answer:

Part 1) The equation of the perpendicular bisector side AB is $y=-\frac{1}{4}x+\frac{15}{4}$

Part 2) The equation of the perpendicular bisector side BC is $y=\frac{2}{3}x+\frac{11}{3}$

Part 3) The equation of the perpendicular bisector side AC is $y=-3x+4$

Part 4) The coordinates of the point P(0.091,3.727)

Step-by-step explanation:

Part 1) Find the equation of the perpendicular bisector side AB

we have

A(–2, 0), B(0, 8)

step 1

Find the slope AB

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{8-0}{0+2}$

$m=4$

step 2

Find the slope of the perpendicular line to side AB

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

therefore

The slope is equal to

$m=-\frac{1}{4}$

step 3

Find the midpoint AB

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M(\frac{-2+0}{2},\frac{0+8}{2})$

$M(-1,4)$

step 4

Find the equation of the perpendicular bisectors of AB

the slope is $m=-\frac{1}{4}$

passes through the point $(-1,4)$

The equation in slope intercept form is equal to

$y=mx+b$

substitute

$4=(-\frac{1}{4})(-1)+b$

solve for b

$b=4-\frac{1}{4}$

$b=\frac{15}{4}$

$y=-\frac{1}{4}x+\frac{15}{4}$

Part 2) Find the equation of the perpendicular bisector side BC

we have

B(0, 8) and C(4, 2)

step 1

Find the slope BC

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{2-8}{4-0}$

$m=-\frac{3}{2}$

step 2

Find the slope of the perpendicular line to side BC

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

therefore

The slope is equal to

$m=\frac{2}{3}$

step 3

Find the midpoint BC

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M(\frac{0+4}{2},\frac{8+2}{2})$

$M(2,5)$

step 4

Find the equation of the perpendicular bisectors of BC

the slope is $m=\frac{2}{3}$

passes through the point $(2,5)$

The equation in slope intercept form is equal to

$y=mx+b$

substitute

$5=(\frac{2}{3})(2)+b$

solve for b

$b=5-\frac{4}{3}$

$b=\frac{11}{3}$

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

Part 3) Find the equation of the perpendicular bisector side AC

we have

A(–2, 0) and C(4, 2)

step 1

Find the slope AC

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{2-0}{4+2}$

$m=\frac{1}{3}$

step 2

Find the slope of the perpendicular line to side AC

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

therefore

The slope is equal to

$m=-3$

step 3

Find the midpoint AC

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M(\frac{-2+4}{2},\frac{0+2}{2})$

$M(1,1)$

step 4

Find the equation of the perpendicular bisectors of AC

the slope is $m=-3$

passes through the point $(1,1)$

The equation in slope intercept form is equal to

$y=mx+b$

substitute

$1=(-3)(1)+b$

solve for b

$b=1+3$

$b=4$

$y=-3x+4$

Part 4) Find the coordinates of the point of concurrency of the perpendicular bisectors (P)

we know that

The point of concurrency of the perpendicular bisectors is called the circumcenter.

Solve by graphing

using a graphing tool

the point of concurrency of the perpendicular bisectors is P(0.091,3.727)

see the attached figure

5 0

3 years ago

Find the square root of each.

DIA [1.3K]

169=13
144=12
Hope this helps and please give brainliest!

8 0

3 years ago