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2 years ago

15=7-|2x| Solve for x in this absolute value equation.

Mathematics

2 answers:

DIA [1.3K]2 years ago

8 0

$15=7-|2x|\\
-|2x|=8\\
|2x|=-8\\
x\in\emptyset$

ivann1987 [24]2 years ago

6 0

$7-|2x|=15\ \ \ \ |subtract\ 7\ from\ both\ sides\\\\-|2x|=8\\\\|2x|=-8-FALSE\\\\x\in\O\ because\ |a|\geq0\ for\ each\ a\in\mathbb{R}$

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2 years ago

At the craft store, Tara bought 12 yards of fabric to make tote bags. Each bag requires 1.25 yards of fabric. If Tara makes 5 to

matrenka [14]

Answer:

5.75 yds left

Step-by-step explanation:

1.25 yds per bag*5 bags =6.25 yds used

She had 12 yds

12-6.25 =5.75

She has 5.75 yds left

5 0

1 year ago

Look at the graph and tell me why it is proportional or nonproportional and why?

andreyandreev [35.5K]

Answer:

I think it is not proportional because the line is supposed to go thro the middle of each dot, and some of the dots don't have a line thro the middle only on the side or not even on the dot.

Hope this helps!

Step-by-step explanation:

please mark me brainliest...

8 0

2 years ago

Write a point slope equation for -5/7 using the coordinates (6,2)

sleet_krkn [62]

Answer:

(13, -3)

Step-by-step explanation:

slope = -5/7

(6,2)

m = ( y2 - y1) / ( x2 - x1 )

-5/7 = ( y2 - 2) / ( x2 - 6 )

-5 = y2 -2

+2 +2

-3 = y2

7 = x2 - 6

+6 +6

13 = x2

Check answer again by log them in:

m = ( y2 - y1 ) / ( x2 - x1 )

m = ( -3 - 2 ) / ( 13 - 6 )

m = -5 / 7

I hope this helps!

7 0

2 years ago

You throw a ball up and its height h can be tracked using the equation h=2x^2-12x+20.

postnew [5]

This problem seems to be wrong because no minimum point was found and no point of landing exists

Answer:

1) There is no maximum height

2) The ball will never land

Step-by-step explanation:

Derivatives

Sometimes we need to find the maximum or minimum value of a function in a given interval. The derivative is a very handy tool for this task. We only have to compute the first derivative f' and have it equal to 0. That will give us the critical points.

Then, compute the second derivative f'' and evaluate the critical points in there. The criteria establish that

If f''(a) is positive, then x=a is a minimum

If f''(a) is negative, then x=a is a maximum

The function provided in the question is

$h(x)=2x^2-12x+20$

Let's find the first derivative

$h'(x)=4x-12$

solving h'=0:

$4x-12=0$

x=3

Computing h''

h''(x)=4

It means that no matter the value of x, the second derivative is always positive, so x=3 is a minimum. The function doesn't have a local maximum or the ball will never reach a maximum height

To find when will the ball land, we set h=0

$2x^2-12x+20=0$