30 pointsssss help meeeee

13. f(x) = x + 5 a. f(4) b. f(7) c. f(-3) d. f(0) e. f(2.4) f. f(2/3)

SashulF [63]

A.9
B.12
C.2
D.5
E.7.4
F.5 2/3 or 5.6666

4 0

3 years ago

100 BRAINLY POINTS!!!

BARSIC [14]

Answer:

C) $y=-0.5x+8$

Step-by-step explanation:

Line of best fit (trendline) : a line through a scatter plot of data points that best expresses the relationship between those points.

All the given options for the line of best fit are linear equations.

Therefore, we can add the line of best fit to the graph (see attached), remembering to have roughly the same number of points above and below the line.

Linear equation: $y=mx+b$

(where $m$ is the slope and $b$ is the y-intercept)

From inspection of the line of best fit, we can see that the y-intercept (where x = 0) is approximately 8. So this suggests that options C or D are the solution.

We can also see that the slope (gradient) of the line of best fit is approximately -0.5 (as the rate of change (y/x) is -1 unit of y for every +2 units of x).

Therefore, C is the solution, and the closet approximation to the line of best fit is $y=-0.5x+8$

4 0

2 years ago

Given that (8,7) is on the graph of f(x), find the corresponding point for the function f(x)+4

ozzi

Answer:

( 8,11)

Step-by-step explanation:

When x = 8 the output is 7

The new function

f(x) +4

when x = 8

The output is f(8) +4= 7+4 = 11

( 8,11)

7 0

3 years ago

Read 2 more answers

Perform the operation and write the answer in simplest form. 637÷11314

Step2247 [10]

Answer:

0.0563

the fraction cannot be simplified

7 0

3 years ago

What are the solutions to the equation

frosja888 [35]

Answer:

C.

$x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$ and $x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i$

Step-by-step explanation:

You have the quadratic function $2x^2-x+1=0$ to find the solutions for this equation we are going to use Bhaskara's Formula.

For the quadratic functions $ax^2+bx+c=0$ with $a\neq 0$ the Bhaskara's Formula is:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}$

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}$

It usually has two solutions.

Then we have $2x^2-x+1=0$ where a=2, b=-1 and c=1. Applying the formula:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}\\\\x_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_1=\frac{1+\sqrt{1-8} }{4}\\\\x_1=\frac{1+\sqrt{-7} }{4}\\\\x_1=\frac{1+\sqrt{(-1).7} }{4}\\x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}$

Observation: $\sqrt{-1}=i$

$x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}\\\\x_1=\frac{1+i.\sqrt{7}}{4}\\\\x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$

And,

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}\\\\x_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_2=\frac{1-i.\sqrt{7} }{4}\\\\x_2=\frac{1}{4}-(\frac{\sqrt{7}}{4})i$

Then the correct answer is option C.

$x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$ and $x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i$

3 0

3 years ago