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3 years ago

Which is a correct first step in solving 5-2x<8x-3

Mathematics

2 answers:

Nady [450]3 years ago

6 0

Add the like terms, 8x and 2x.
I hope that helped! :)

alexandr1967 [171]3 years ago

3 0

Answer:

first step is combining like terms, subtract 8x from both sides

Step-by-step explanation:

$5-2x$

To solve any inequality , move all x terms together

To remove 8x from the right side, subtract 8x from both sides

$5-10x$

So first step is combining like terms

Subtract 5 from both sides

$-10x$

Divide both sides by -10

$x>\frac{4}{5}$

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Simplify 8 over negative 4 divided by negative 3 over 9. 6 −6 12 −12

skelet666 [1.2K]

Answer:

Step-by-step explanation:

Given

$\frac{8}{-4} / \frac{-3}{9}$

Required

Solve

$\frac{8}{-4} / \frac{-3}{9}$

We start by converting the division sign to multiplication

$\frac{8}{-4} / \frac{-3}{9} =\frac{8}{-4} * \frac{-9}{3}$

Then, we divide -9 by 3

$\frac{8}{-4} / \frac{-3}{9}=\frac{8}{-4} * -3$

Then, we divide 8 by -4

$\frac{8}{-4} / \frac{-3}{9}=-2 * -3$

Multiply -2 and -3 to give final result

$\frac{8}{-4} / \frac{-3}{9} = 6$

Hence, $\frac{8}{-4} / \frac{-3}{9}$ is equivalent to 6

4 0

3 years ago

Angelo's mother put $50 on his lunch card. He spends the same amount

amid [387]

Answer:

Given that Angelo spends the same amount every day from the amount in

the lunch card, the function of the amount remaining is a linear function.

The constant rate of change of the function is; -5.25

The two ordered pairs used to find the constant rate of change are; (1, 44.75) and (2, 39.5)

Reasons:

The amount Angelo's mother put on the lunch card = $50

A possible table of values to the question is presented as follows;

Required:

The constant rate of the function that gives the amount remaining from the

amount Angelo's mother put on his lunch card.

Solution:

The two ordered pairs that can be used to find the slope or constant rate of change are;

(x₁, y₁) = (1, 44.75), and (x₂, y₂) = (2, 39.5)

With the above two ordered pairs, we have the constant rate of change of the function given as follows;

The constant rate of change for the function that gives the amount remaining in the lunch card is; -5.25

3 0

3 years ago

Suppose two parallel lines are cut by a transversal.which angle must be supplementary

Alborosie

Adjacent angles are supplementary

4 0

3 years ago

Find the distance between lines 8x−15y+5=0 and 16x−30y−12=0.

AlladinOne [14]

Answer:

The distance between the two parallel lines is 11/17 units

Step-by-step explanation:

we have

8x-15y+5=0 -----> equation A

16x-30y-12=0 ----> equation B

Divide by 2 both sides equation B

8x-15y-6=0 ----> equation C

Compare equation A and equation C

Line A and Line C are parallel lines with different y-intercept

step 1

Find the slope of the parallel lines (The slope of two parallel lines is the same)

8x-15y+5=0

15y=8x+5

$y=\frac{8}{15}x+\frac{1}{3}$

the slope is

$m=\frac{8}{15}$

step 2

Find the slope of a line perpendicular to the given lines

Remember that

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

$m1*m2=-1$

we have

$m1=\frac{8}{15}$

therefore

$m2=-\frac{15}{8}$

step 3

Find the equation of the line perpendicular to the given lines

assume any point that lie on line A

$y=\frac{8}{15}x+\frac{1}{3}$

For $x=0$

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

To find the equation of the line we have

$point\ (0,1/3)$ ---> is the y-intercept

$m=-\frac{15}{8}$

The equation in slope intercept form is

$y=-\frac{15}{8}x+\frac{1}{3}$ -----> equation D

step 4

Find the intersection point of the perpendicular line with the Line C

we have the system of equations

$y=-\frac{15}{8}x+\frac{1}{3}$ ----> equation D

$8x-15y-6=0$ ----> $y=\frac{8}{15}x-\frac{2}{5}$ ----> equation E

equate equation D and equation E and solve for x

$\frac{8}{15}x-\frac{2}{5}=-\frac{15}{8}x+\frac{1}{3}$

$\frac{8}{15}x+\frac{15}{8}x=\frac{1}{3}+\frac{2}{5}$

Multiply by 120 both sides to remove fractions

$64x+225x=40+48$

$289x=88$

$x=88/289$

Find the value of y

$y=-\frac{15}{8}(88/289)+\frac{1}{3}$

$y=-\frac{206}{867}$

the intersection point is $(\frac{88}{289},-\frac{206}{867})$

step 5

Find the distance between the points $(0,\frac{1}{3})$ and $(\frac{88}{289},-\frac{206}{867})$

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

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

substitute the values

$d=\sqrt{(-\frac{206}{867}-\frac{1}{3})^{2}+(\frac{88}{289}-0)^{2}}$

$d=\sqrt{(-\frac{495}{867})^{2}+(\frac{88}{289})^{2}}$

$d=\sqrt{(\frac{245,025}{751,689})+(\frac{7,744}{83,521})}$

$d=\sqrt{\frac{314,721}{751,689}}$

$d=\frac{561}{867}\ units$

Simplify

$d=\frac{11}{17}\ units$

therefore

The distance between the two parallel lines is 11/17 units

see the attached figure to better understand the problem

4 0

3 years ago

. Write a word problem that can be acted out using these counters.

Rufina [12.5K]

Please comment on this answer and give me a full question and I will be just fine answering it for you.

8 0

3 years ago