All categories

3 years ago

Only answer question 28 please lots of points

Mathematics

1 answer:

Blababa [14]3 years ago

4 0

The line plot smallest number always goes first biggest number goes last

You might be interested in

Michelle borrowed $10,125 at 12.5 percent for 10 years. What is her monthly payment

mash [69]

Answer:

$148.21

Step-by-step explanation:

A suitable financial calculator, web site, or spreadsheet can figure this for you. Or you can use the formula given in your reference material (text or web site).

7 0

3 years ago

Read 2 more answers

What is the mean of -71,-62,-88-38-89,15,-76

olga nikolaevna [1]

71
62
88
38
89
76
-15 (i'm inversing everything to make it simple)
The mean is -69.28

6 0

3 years ago

HELP PLZ

ollegr [7]

Answer:

a. ∠D'A'B'

Step-by-step explanation:

From the graph it is clear that the transformation doesn't affect the size and shape of the figure ABCD.

Only the coordinates of the points are changed with the length of each side remaining the same.

Hence the corresponding angle values will also remain unaltered.

∴ ∠DAB = ∠D'A'B'

5 0

3 years ago

| x | >= 4 draw the graph

-Dominant- [34]

Answer:

Solve for x

x ≤ −4 or x ≥ 4

6 0

3 years ago

g Use this to find the equation of the tangent line to the parabola y = 2 x 2 − 7 x + 6 at the point ( 4 , 10 ) . The equation o

natali 33 [55]

Answer:

The tangent line to the given curve at the given point is $y=9x-26$ .

Step-by-step explanation:

To find the slope of the tangent line we to compute the derivative of $y=2x^2-7x+6$ and then evaluate it for $x=4$ .

$(y=2x^2-7x+6)'$ Differentiate the equation.

$(y)'=(2x^2-7x+6)'$ Differentiate both sides.

$y'=(2x^2)'-(7x)'+(6)'$ Sum/Difference rule applied: $(f(x)\pmg(x))'=f'(x)\pm g'(x)$

$y'=2(x^2)'-7(x)'+(6)'$ Constant multiple rule applied: $(cf)'=c(f)'$

$y'2(2x)-7(1)+(6)'$ Applied power rule: $(x^n)'=nx^{n-1}$

$y'=4x-7+0$ Simplifying and apply constant rule: $(c)'=0$

$y'=4x-7$ Simplify.

Evaluate y' for x=4:

$y'=4(4)-7$

$y'=16-7$

$y'=9$ is the slope of the tangent line.

Point slope form of a line is:

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

where $m$ is the slope and $(x_1,y_1)$ is a point on the line.

Insert 9 for $m$ and (4,10) for $(x_1,y_1)$ :

$y-10=9(x-4)$

The intended form is $y=mx+b$ which means we are going need to distribute and solve for $y$ .

Distribute:

$y-10=9x-36$

Add 10 on both sides:

$y=9x-26$

The tangent line to the given curve at the given point is $y=9x-26$ .

------------Formal Definition of Derivative----------------

The following limit will give us the derivative of the function $f(x)=2x^2-7x+6$ at $x=4$ (the slope of the tangent line at $x=4$ ):

$\lim_{x \rightarrow 4}\frac{f(x)-f(4)}{x-4}$

$\lim_{x \rightarrow 4}\frac{2x^2-7x+6-10}{x-4}$ We are given f(4)=10.

$\lim_{x \rightarrow 4}\frac{2x^2-7x-4}{x-4}$

Let's see if we can factor the top so we can cancel a pair of common factors from top and bottom to get rid of the x-4 on bottom:

$2x^2-7x-4=(x-4)(2x+1)$

Let's check this with FOIL:

First: $x(2x)=2x^2$

Outer: $x(1)=x$

Inner: $(-4)(2x)=-8x$

Last: $-4(1)=-4$

---------------------------------Add!

$2x^2-7x-4$

So the numerator and the denominator do contain a common factor.

This means we have this so far in the simplifying of the above limit:

$\lim_{x \rightarrow 4}\frac{2x^2-7x-4}{x-4}$

$\lim_{x \rightarrow 4}\frac{(x-4)(2x+1)}{x-4}$

$\lim_{x \rightarrow 4}(2x+1)$

Now we get to replace x with 4 since we have no division by 0 to worry about:

2(4)+1=8+1=9.

6 0

3 years ago