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poizon [28]
3 years ago
10

X squared = −169. A= undefined B= 13 C= ±13 D= −13

Mathematics
1 answer:
mash [69]3 years ago
7 0

Answer:

A.   Undefined.

Step-by-step explanation:

There is no real square root of a negative number.

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What is the slope of the line that passes through the points ( 5 , 6 )and ( 5 , 9 ) ? Write your answer in simplest form.
Lapatulllka [165]

Answer:

3

Step-by-step explanation:

9-6

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m=3

8 0
2 years ago
**Spam answers will not be tolerated**
Morgarella [4.7K]

Answer:

f'(x)=-\frac{2}{x^\frac{3}{2}}

Step-by-step explanation:

So we have the function:

f(x)=\frac{4}{\sqrt x}

And we want to find the derivative using the limit process.

The definition of a derivative as a limit is:

\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

Therefore, our derivative would be:

\lim_{h \to 0}\frac{\frac{4}{\sqrt{x+h}}-\frac{4}{\sqrt x}}{h}

First of all, let's factor out a 4 from the numerator and place it in front of our limit:

=\lim_{h \to 0}\frac{4(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x})}{h}

Place the 4 in front:

=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}

Now, let's multiply everything by (√(x+h)(√(x))) to get rid of the fractions in the denominator. Therefore:

=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}(\frac{\sqrt{x+h}\sqrt x}{\sqrt{x+h}\sqrt x})

Distribute:

=4\lim_{h \to 0}\frac{({\sqrt{x+h}\sqrt x})\frac{1}{\sqrt{x+h}}-(\sqrt{x+h}\sqrt x)\frac{1}{\sqrt x}}{h({\sqrt{x+h}\sqrt x})}

Simplify: For the first term on the left, the √(x+h) cancels. For the term on the right, the (√(x)) cancel. Thus:

=4 \lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }

Now, multiply both sides by the conjugate of the numerator. In other words, multiply by (√x + √(x+h)). Thus:

= 4\lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }(\frac{\sqrt x +\sqrt{x+h})}{\sqrt x +\sqrt{x+h})}

The numerator will use the difference of two squares. Thus:

=4 \lim_{h \to 0} \frac{x-(x+h)}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}

Simplify the numerator:

=4 \lim_{h \to 0} \frac{x-x-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}\\=4 \lim_{h \to 0} \frac{-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}

Both the numerator and denominator have a h. Cancel them:

=4 \lim_{h \to 0} \frac{-1}{(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}

Now, substitute 0 for h. So:

=4 ( \frac{-1}{(\sqrt{x+0}\sqrt x)(\sqrt x+\sqrt{x+0})})

Simplify:

=4( \frac{-1}{(\sqrt{x}\sqrt x)(\sqrt x+\sqrt{x})})

(√x)(√x) is just x. (√x)+(√x) is just 2(√x). Therefore:

=4( \frac{-1}{(x)(2\sqrt{x})})

Multiply across:

= \frac{-4}{(2x\sqrt{x})}

Reduce. Change √x to x^(1/2). So:

=-\frac{2}{x(x^{\frac{1}{2}})}

Add the exponents:

=-\frac{2}{x^\frac{3}{2}}

And we're done!

f(x)=\frac{4}{\sqrt x}\\f'(x)=-\frac{2}{x^\frac{3}{2}}

5 0
2 years ago
A box is placed on the floor.
Sedaia [141]
Hiiii hope it helps
8 0
2 years ago
The ball followed a path modelled by the equation h = −0.001! + 0.5 + 2.5 where h is the height of the ball in feet and is the h
Mama L [17]

The heights the balls hit a fence at 350 ft distance are 65 feet, 38 feet and 30 feet, respectively

<h3>Represent the distance-height relationship for each player’s ball as an equation, in a table and on a graph. </h3>

<u>Juan</u>

Juan's equation is given as:

h = -0.001d^2 + 0.5d + 2.5

h =

Set d to multiples of 50 from 0 to 400.

So, the table of values of Juan's function is:

d (ft)                   h(ft)

0                          2.5

50                        25

100                      42.5

150                        55

200                      62.5

250                        65

300                      62.5

350                        65

400                      42.5

See attachment for the graph of Juan's function

<u>Mark</u>

A quadratic function is represented as:

h = ad^2 + bd + c

Using the values on the table of values, we have:

c = 3 -- the constant value

So, the equation becomes

h = ad^2 + bd + 3

Using the two other values on the table of values, we have:

23 = a(50)^2 + b(50) + 3

38 = a(100)^2 + b(100) + 3

Using a graphing tool, we have:

a = -0.001

b = 0.45

So, Mark's equation is h(d) = -0.001d^2 + 0.45d + 3

See attachment for Mark's graph.

<u>Barry</u>

From the graph, we have the table of values of Barry's function to be:

d (ft)                   h(ft)

0                          2.5

50                        21

100                      35

150                       44

200                      48

250                       46

300                      41

350                       30

400                      14

450                      0

Using a graphing tool, we have the quadratic function to be:

y = -0.001x^2 +0.4x +2.5

<h3><u>The shortest and the greatest distance before hitting the ground</u></h3>

From the graphs, equations and tables, the distance travelled by the balls are:

Juan = 505 feet

Mark = 457 feet

Barry = 450 feet

This means that Juan's ball would travel the greatest distance while Barry's ball would travel the shortest.

<h3>The height the balls hit a fence at 350 ft distance</h3>

To do this, we set d = 350

From the graphs, equations and tables, the height at 350 ft by the balls are:

Juan = 65 feet

Mark = 38 feet

Barry = 30 feet

The above represents the height the balls hit the fence

Read more about quadratic functions at:

brainly.com/question/12446886

#SPJ1

4 0
1 year ago
Please help me 7th grade
Anon25 [30]

Answer:

4(-80) = -320

Step-by-step explanation:

His elevation decreased by 80 each hour. Since he was walking for 4 hours, the equation would be

4(-80) = -320.

6 0
2 years ago
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