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3 years ago

Which of the following equations corresponds to the graph below?

Mathematics

2 answers:

mrs_skeptik [129]3 years ago

6 0

Answer: D: Y= -2x - 1

Step-by-step explanation:

y= mx+b

m= slope

b= y intercept

notka56 [123]3 years ago

6 0

Answer:

Step-by-step explanation:

Find slope.

rise over run

It goes up two and back one, so the slope is -2.

y int is at -1. plug it into y = mx + b

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The table is shows the fraction of the votes that each candidate received. If 230 students voted, how many students voted for ea

barxatty [35]

3/5 * 230 = 138 votes for Nyemi

3/10 * 230 = 69 votes for Luke

1/10 * 230 = 23 votes for Natalie

***************** DOUBLE-CHECK ****************

138 + 69 + 23 = 230 Correct!!

6 0

2 years ago

Read 2 more answers

Finding an Equation of a tangent Line in Exercise, find an equation of the tangent line to the graph of the function at the give

frutty [35]

Answer:

$y=\dfrac{3x}{e}+\dfrac{4}{e}$

this is the equation of the tangent at point (-1,1/e)

Step-by-step explanation:

to find the tangent line we need to find the derivative of the function g(x).

$g(x) =e^{x^3}$

we know that $\frac{d}{dx}(e^{f(x)})=e^{f(x)}f'(x)$

$g'(x) =e^{x^{3}}(3 x^{2})$

$g'(x) =3 x^{2} e^{x^{3}}$

this the equation of the slope of the curve at any point x and it also the slope of the tangent at any point x. hence, g'(x) can be denoted as 'm'

to find the slope at (-1,1/e) we'll use the x-coordinate of the point i.e. x = -1

$m =3 (-1)^{2} e^{(-1)^{3}}\\m =3e^{-1}\\m=\dfrac{3}{e}$

using the equation of line:

$(y-y_1)=m(x-x_1)$

we'll find the equation of the tangent line.

here (x1,y1) =(-1,1/e), and m = 3/e

$(y-\dfrac{1}{e})=\dfrac{3}{e}(x+1)\\y=\dfrac{3x}{e}+\dfrac{3}{e}+\dfrac{1}{e}\\$

$y=\dfrac{3x}{e}+\dfrac{4}{e}$

this is the equation of the tangent at point (-1,1/e)

3 0

2 years ago

0=4t-16t^2 Solve this please

vaieri [72.5K]

Answer:

$\large\boxed{t=0\ \vee\ t=\dfrac{1}{4}}$

Step-by-step explanation:

$4t-16t^2=0\qquad\text{divide both sides by 4}\\\\\dfrac{4t}{4}-\dfrac{16t^2}{4}=\dfrac{0}{4}\\\\t-4t^2=0\\\\t(1-4t)=0\iff t=0\ \vee\ 1-4t=0\\\\1-4t=0\qquad\text{subtract 1 from both sides}\\\\-4t=-1\qquad\text{divide both sides by (-4)}\\\\t=\dfrac{1}{4}$