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3 years ago

Help fast!!!!!!!!!!!!! It's due at 12:00 and it's 11:23 please help me i need to get to 90

Mathematics

2 answers:

ryzh [129]3 years ago

8 0

Answer:

slope = $\frac{5}{3}$

Step-by-step explanation:

Select two points (or 2 sets of coordinates) on the graph: (4, 0) and (7, 5).

$X_{1} = 4\\Y_{1} = 0\\X_{2} = 7\\Y_{2} = 5$

Using the slope equation:

m = $(\frac{Y_{2} - Y_{1}}{X_{2} - X_{1}} )$

= $\frac{(5 - 0)}{(7 - 4)}\\$

m = $\frac{5}{3}$

katrin2010 [14]3 years ago

5 0

You can use two method

First method: form a triangle from x = 4 to x = 7. See vertically and horizontally (count the squares). It would be 5 and 3. Therefore 5/3 would be the slope.

Second method: use the slope formula

Determine 2 points on the graph

(4,0) and (7,5)

Formula: (y2-y1)/(x2-x1)

(5-0)/(7-4) = 5/3

The slope is 5/3

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Natali5045456 [20]

Answer:

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Step-by-step explanation:

Given that , the sum of the first nine terms of an arithmetic series is 162 and the sum of the first 12 terms is 288.

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8 0

3 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6) Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

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3 years ago

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kobusy [5.1K]

Answer:

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Step-by-step explanation:

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4 years ago

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To find a percent, you use this simple formula:
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3 years ago