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3 years ago

Which equation is a point-slope form of the equation of this line (-2, 1 ) and (1, 7 )

Mathematics

1 answer:

lana [24]3 years ago

6 0

The point-slope formula:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

We have (-2, 1) and (1, 7).

Substitute:

$m=\dfrac{7-1}{1-(-2)}=\dfrac{6}{3}=2$

$y-1=2(x-(-2))\\\\y-1=2(x+2)$

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Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^

Gemiola [76]

Answer:

$\rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x}$

Step-by-step explanation:

we would like to figure out the differential coefficient of $e^{2x}(1+\ln(x))$

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

$\displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )$

to do so distribute:

$\displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x}$

take derivative in both sides which yields:

$\displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )$

by sum derivation rule we acquire:

$\rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x}$

Part-A: differentiating $e^{2x}$

$\displaystyle \frac{d}{dx} {e}^{2x}$

the rule of composite function derivation is given by:

$\rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)$

so let g(x) [2x] be u and transform it:

$\displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x$

differentiate:

$\displaystyle {e}^{u} \cdot 2$

substitute back:

$\displaystyle \boxed{2{e}^{2x} }$

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

$\displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)$

let

$f(x) \implies \ln(x)$
$g(x) \implies {e}^{2x}$

substitute

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x}$

differentiate:

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }$

Final part:

substitute what we got:

$\rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }$

and we're done!

6 0

3 years ago

The temperature is -2 degree Fearing height if it drops 1 degree every 10 minutes what will the temperature be in an hour 1 hour

ExtremeBDS [4]

-2
1 degree ever 10mins
1 hour = 60mins
1x6
Drops 6°f
-2+-6 = -8 degrees after 1 hours

8 0

3 years ago

15^2x=36 write the equation in logarithmic form.

marusya05 [52]

X=log36/(2*log15)
X=aprox 0.6616

5 0

3 years ago

Which represents the solution(s) of the system of equations, y = x2 – 2x – 15 and y = 8x – 40? Determine the solution set algebr

almond37 [142]

Answer:

(5, 0 )

Step-by-step explanation:

Given the 2 equations

y = x² - 2x - 15 → (1)

y = 8x - 40 → (2)

Substitute y = x² - 2x - 15 into (2)

x² - 2x - 15 = 8x - 40 ( subtract 8x - 40 from both sides )

x² - 10x + 25 = 0 ← in standard form

(x - 5)² = 0 ← in factored form, thus

x - 5 = 0 ⇒ x = 5

Substitute x = 5 into (2) for corresponding value of y

y = 8(5) - 40 = 40 - 40 = 0

Solution is (5, 0 )

3 0

3 years ago

Please help! Its due today!

Helga [31]

Answer:

don't know if this is the right way butttt i'm going to give it a shot

Step-by-step explanation:

Cost: I'm guessing this is for the rent for the shoes or is the total cost?

Bowler World: 5 + 1.1g = c | c + 1.1g = t {(this is the other option) probs not correct}

Lucky Spares: 3 + 1.5g = c | c + 1.5g = t

K so personally you should choose option 1 (for example: 5 + 1.1g = c), i have no idea how the proper format is supposed to be

either i hope this helps :)

7 0

2 years ago