Ask question

Ask question

All categories

goldfiish [28.3K]

3 years ago

9

How to write an answer in Brainly with decorative style like this

1 answer:

Marizza181 [45]3 years ago

7 0

Answer: Go to ur profile and choose the tab that says which questions u have asked then you can edit ur response (a little pencil icon will be on the top right corner of ur question)

Step-by-step explanation:

info above.

You might be interested in

Record your<br> ! 1)<br> 5x = 125

KengaRu [80]

The answer is X= 25
You just have to divide 125 by 5

3 0

3 years ago

Three friends earned more than $200 washing cars they pay their parents $20 for supplies and divided the rest of the money equal

kow [346]

$90 because 200-10=180 and divide 180 by 2 which is 90.

3 0

3 years ago

Read 2 more answers

Write the equation for Chelsea's graph in

xeze [42]

Answer:

$y=-4x+16$ .

Step-by-step explanation:

From the given graph it is clear that the y-intercept is 16.

The line passes through (-2,24) and (2,8). So, slope of line is

$Slope=\dfrac{y_2-y_1}{x_2-x_1}$

$Slope=\dfrac{8-24}{2-(-2)}$

$Slope=\dfrac{-16}{4}$

$Slope=-4$

Slope intercept form of a line is

$y=mx+b$

where, m is slope and b is y-intercept.

Substitute m=-4 and b=16 in the above equation.

$y=(-4)x+(16)$

$y=-4x+16$

Therefore, the required equation is $y=-4x+16$ .

6 0

3 years ago

In Exercises 6 and 7, use the given measure to find the missing value in each data set. 6. The mean of the ages of 5 brothers is

strojnjashka [21]

Solution

Problem 6

For this case we can do this:

12, 16,__, 14, 8, 7

We can solve for x like this:

$\frac{12+16+x+14+8+7}{6}=13$ $x=13\cdot6-(12+16+14+8+7)=21$

Problem 7

F

7 0

1 year ago

Limit definition for slope of the graph, equation of tangent line point for<br> f(x)=2x^2 at x=(-1)

Tems11 [23]

The slope of the tangent line to $f$ at $x=-1$ is given by the derivative of $f$ at that point:

$f'(-1)=\displaystyle\lim_{x\to-1}\frac{f(x)-f(-1)}{x-(-1)}=\lim_{x\to-1}\frac{2x^2-2}{x+1}$

Factorize the numerator:

$2x^2-2=2(x^2-1)=2(x-1)(x+1)$

We have $x$ approaching -1; in particular, this means $x\neq-1$ , so that

$\dfrac{2x^2-2}{x+1}=\dfrac{2(x-1)(x+1)}{x+1}=2(x-1)$

Then

$f'(-1)=\displaystyle\lim_{x\to-1}\frac{2x^2-2}{x+1}=\lim_{x\to-1}2(x-1)=2(-1-1)=-4$

and the tangent line's equation is

$y-f(-1)=f'(-1)(x-(-1))\implies y-4x-2$

6 0

3 years ago