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3 years ago

-16x-10>14-10x hi can u please help me with this

Mathematics

2 answers:

kolbaska11 [484]3 years ago

7 0

Answer:

Step-by-step explanation:

Let's solve your inequality step-by-step.

−16x−10>14−10x

Step 1: Simplify both sides of the inequality.

−16x−10>−10x+14

Step 2: Add 10x to both sides.

−16x−10+10x>−10x+14+10x

−6x−10>14

Step 3: Add 10 to both sides.

−6x−10+10>14+10

−6x>24

Step 4: Divide both sides by -6.

−6x

−6

x<−4

Answer:

x<−4

kompoz [17]3 years ago

4 0

Answer:

x < -4

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Step-by-step explanation:

Step 1: Define inequality

-16x - 10 > 14 - 10x

Step 2: Solve for x

Add 16x to both sides: -10 > 14 + 6x
Subtract 14 on both sides: -24 > 6x
Divide 6 on both sides: -4 > x
Rewrite: x < -4

Here we see that any value x that is less than -4 would work as a solution.

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What is the area of a parallelogram whose vertices are A(−12, 2) , B(6, 2) , C(−2, −3) , and D(−20, −3) ? Enter your answer in t

nikdorinn [45]

Let

$A(-12,2)\\B(6,2)\\C(-2,-3) \\D(-20.-3)\\E(-12.-3)$

using a graphing tool

see the attached figure to better understand the problem

we know that

Parallelogram is a quadrilateral with opposite sides parallel and equal in length

$AB=CD \\AD=BC$

The area of a parallelogram is equal to

$A=B*h$

where

B is the base

h is the height

the base B is equal to the distance AB

the height h is equal to the distance AE

Step 1

Find the distance AB

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

$A(-12,2)\\B(6,2)$

substitute the values

$d=\sqrt{(2-2)^{2}+(6+12)^{2}}$

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Step 2

Find the distance AE

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

$A(-12,2)\\E(-12.-3)$

substitute the values

$d=\sqrt{(-3-2)^{2}+(-12+12)^{2}}$

$d=\sqrt{(-5)^{2}+(0)^{2}}$

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Step 3

Find the area of the parallelogram

The area of a parallelogram is equal to

$A=B*h$

$A=AB*AE$

substitute the values

$A=18*5=90\ units^{2}$

therefore

the answer is

the area of the parallelogram is $90\ units^{2}$

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