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natita [175]
3 years ago
11

Write the linear equation in slope intercept form passing through the point with the given slope points (-8,6) slope =1/5

Mathematics
1 answer:
Arte-miy333 [17]3 years ago
5 0
Y = 1/5x + 7.6
6 = 1/5(-8) + 7.6
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Look at the table to answer the question.
Alexus [3.1K]

Answer:

Your power function is  y=-2x^{2}

Step-by-step explanation:

The way I found it was.

0=0,  so, the function multiply by zero, and have no other term to add

A number (one in a case) gives as a result: -2.  So, one, elevated to ANY power results in one, every time. So, I have 1, as a factor and -2 as a result. One is as well a factor that delivers a result equal to the other factor. (3)(1)=3 , (8987)(1)=8987. So, the other factor must be -2

Then I checked all the table, and the results were consistent.


5 0
3 years ago
Challenge Question Sally's bank account balance was $50 on Monday. On Tuesday, she bought a new pair of shoes for $45. On Thursd
Alex17521 [72]

Answer:

  $8

Step-by-step explanation:

After the shoe purchase, her balance was $50 -45 = $5.

After depositing her babysitting earnings, her balance was $5+25 = $30.

After buying pizza, her balance was $30 -42 = -$12.

<em>Assuming no overdraft charge</em>, her balance after depositing her earnings was ...

  -$12 +20 = $8

Sally's balance on Sunday was $8.

7 0
2 years ago
Gordon types 1,922 words in 31 minutes. Find the unit rate
Kisachek [45]

Answer:

64.2580645161 words in a minute

Step-by-step explanation:

Hope that helps!

8 0
2 years ago
Read 2 more answers
Graph a line that contains the point (3,-6) and has a slope of -1/2
djyliett [7]
If I were you, I would make the starting point (3,-6). From there, you will want to use the slope of -1/2 (go down 1 unit and to the right 2 units and draw a point)
4 0
3 years ago
1. (5pts) Find the derivatives of the function using the definition of derivative.
andreyandreev [35.5K]

2.8.1

f(x) = \dfrac4{\sqrt{3-x}}

By definition of the derivative,

f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}

We have

f(x+h) = \dfrac4{\sqrt{3-(x+h)}}

and

f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}

Combine these fractions into one with a common denominator:

f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}

3.1.1.

f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3

Differentiate one term at a time:

• power rule

\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4

\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}

\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}

The last two terms are constant, so their derivatives are both zero.

So you end up with

f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}

8 0
2 years ago
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