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3 years ago

PLEASE HELP ME!!!!!

Mathematics

1 answer:

iVinArrow [24]3 years ago

4 0

Answer:

3.92%

Step-by-step explanation:

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Mar 10, 11:28:13 PM

Virty [35]

Answer:

s249

Step-by-step explanation:

i dont have an explanation but yeah sorry

5 0

3 years ago

Read 2 more answers

Please Help, I'm stuck on this question .

snow_lady [41]

Answer:

24y^15 - 24y^10 + 7

Step-by-step explanation:

f(2y^5)

= 3(2y^5)^3 - 6(2y^5)^2 +7

= 3(8y^15) - 6(4y^10) + 7

= 24y^15 - 24y^10 + 7

4 0

3 years ago

And the first derivative of<br>f(n) = 15e

kirill [66]

Answer:

the answer is 0.

Step-by-step explanation:

The answer is 0, I had a question like this before

3 0

3 years ago

Read 2 more answers

Write in slope intercept form an equation of the line that passes through (-2,5),(-1,1)

Juliette [100K]

Answer:

y=-4x-3

Step-by-step explanation:

Hi there!

We are given the following two points: (-2,5) and (-1,1)

We want to write the equation of the line that passes through these two lines in slope-intercept form

Slope-intercept form is given as y=mx+b, where m is the slope and b is the y intercept

First, let's find the slope

The formula for the slope (m) calculated from 2 points is $\frac{y_2-y_1}{x_2-x_1}$ , where $(x_1, y_1)$ and $(x_2, y_2)$ are points

We have everything we need to calculate the slope, but let's label the value of the points to avoid any confusion

$x_1=-2\\y_1=5\\x_2=-1\\y_2=1$

Now substitute into the formula

m= $\frac{y_2-y_1}{x_2-x_1}$

m= $\frac{1-5}{-1--2}$
Simplify

m= $\frac{1-5}{-1+2}$

Subtract

m= $\frac{-4}{1}$

Divide

m=-4

Now substitute -4 as m into the equation:

y=-4x+b

Now we need to find b

As the equation passes through both (-2, 5) and (-1, 1), we can use either point to help solve for b.

Taking (-2,5) for example:

Substitute -2 as x and 5 as y:

5=-4(-2)+b

Multiply

5=8+b

Subtract 8 from both sides

-3=b

Substitute -3 as b.

y=-4x-3

Hope this helps!

See more on this topic here: brainly.com/question/27304092

4 0

2 years ago

Find the limit of the function algebraically. limit as x approaches negative three of quantity x squared minus nine divided by q

iris [78.8K]

The limand is continuous at $x=-3$ , so you can directly substitute $x=-3$ to get

$\displaystyle\lim_{x\to-3}\frac{x^2-9}{x^3+3}=\frac{(-3)^2-9}{(-3)^3+3}=\dfrac{9-9}{-27+3}=0$

Did you mean to write $x^3+3^3=x^3+27$ in the denominator by any chance? In that case, you would instead have

$\displaystyle\lim_{x\to-3}\frac{x^2-9}{x^3+3}=\lim_{x\to-3}\frac{(x+3)(x-3)}{(x+3)(x^2-3x+9)}=\lim_{x\to-3}\frac{x-3}{x^2-3x+9}=\frac{-3-3}{(-3)^2-3(-3)+9}=-\frac29$

5 0

4 years ago