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3 years ago

Solve for x: −4|x + 5| = −16

Mathematics

2 answers:

san4es73 [151]3 years ago

7 0

-4 |x+5| = -16

Divide both sides by -4

|x+5| = 4

x = -1

wel3 years ago

7 0

$-4|x + 5| = -16\\\\|x+5|=4\\\\x+5=4 \vee x+5=-4\\\\x=-1 \vee x=-9$

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4. Which of the following lines is perpendicular to y = -2X + 8?

Alenkinab [10]

For this case we have to by definition, if two lines are perpendicular then the product of its slopes is -1.

That is to say:

$m_ {1} * m_ {2} = - 1$

We have the following equation:

$y = -2x + 8$

So:

$m_ {1} = - 2$

Thus:

$m_ {2} = \frac {-1} {m_ {1}}\\m_ {2} = \frac {-1} {- 2}\\m = \frac {1}{2}$

Thus, a line perpendicular to the given line must have slope $m = \frac {1} {2}.$

Option A:

$x + 2y = 8\\2y = -x + 8\\y = - \frac {1} {2} x + 4$

It is not perpendicular!

Option B:

$x-2y = 6\\2y = x-6\\y = \frac {1} {2} x-3$

If it is perpendicular!

Option C:

$2x + y = 4\\y = -2x + 4$

It is not perpendicular!

Option D:

$2x-y = 1\\y = 2x-1$

It is not perpendicular!

The correct option is option B

ANswer:

Option B

5 0

3 years ago

MaryAnne has 90 quilt squares. She plans to use 32 of them to make a quilt. What percent of her quilt square stock is she using?

Gemiola [76]

Percentage = ( number used / total number) x 100

32/90 = 0.3555

0.3555 x 100 = 35.56%

Rounded to nearest tenth = 35.6%

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2 years ago

A car and a motorcycle leave at noon from the same location, heading in the same direction. The average speed of the car is 30 m

Nonamiya [84]

The car is going at a speed of 50 mph and the motorcycle is going at a speed of 40 mph.
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3 years ago

Read 2 more answers

What is the value of \dfrac{d}{dx}\left(\dfrac{2x+3}{3x^2-4}\right) dx d ( 3x 2 −4 2x+3 )start fraction, d, divided by, d, x

stellarik [79]

Answer:

Step-by-step explanation:

We are asked to find the value of expression $\frac{d}{dx}(\frac{2x+3}{3x^2-4})$ at $x=-1$ .

First of all, we will find the derivative of the given expression using "Quotient Rule of Derivatives" as shown below:

$(\frac{f(x)}{g(x)})'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{(g(x))^2}$

$\frac{d}{dx}(\frac{2x+3}{3x^2-4})$

$\frac{\frac{d}{dx}(2x+3)*(3x^2-4)-(2x+3)*\frac{d}{dx}(3x^2-4)}{(3x^2-4)^2}$

$\frac{2*(3x^2-4)-(2x+3)*(6x)}{(3x^2-4)^2}$

$\frac{6x^2-8-12x^2-18x}{(3x^2-4)^2}$

$\frac{-6x^2-18x-8}{(3x^2-4)^2}$

Therefore, our required derivative is $\frac{-6x^2-18x-8}{(3x^2-4)^2}$ .

Now, we will substitute $x=-1$ in our derivative to find the required value as:

$\frac{-6(-1)^2-18(-1)-8}{(3(-1)^2-4)^2}$

$\frac{-6(1)+18-8}{(3(1)-4)^2}$

$\frac{-6+18-8}{(3-4)^2}$

$\frac{4}{(-1)^2}$

$\frac{4}{1}$

$4$

Therefore, the value of expression $\frac{d}{dx}(\frac{2x+3}{3x^2-4})$ at $x=-1$ is 4.

6 0

3 years ago

PLS HELP ILL GIVE BRAINLIEST

bija089 [108]

Answer:

g(2) = -1

Step-by-step explanation:

When x = 2, we can pick the last expression in the g(x) limit, since our x-value matches the criteria of that expression (our x value is equal to 2)

Now we just plug in our x-value and solve for g(x) [in this case, g(2)]

g(2) = 2x - 5

g(2) = 2(2) - 5

g(2) = 4 - 5

g(2) = -1

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2 years ago