PLEASE HELP!!!! Algebra 1 Which function contains the points shown on the graph?

Find the slope of the line passing through the points -2,7 and -6,3

Helga [31]

Answer: the slope is 1

Step-by-step explanation:

m = y2 - y1 / x2 - x1

m = 7-3 / -2+6

m = 4/4

m=1

6 0

2 years ago

How do you find if a graph is linear

aleksandr82 [10.1K]

To find out of a graph is linear you have to use the vertical line test.

3 0

3 years ago

The braking distance, in feet of a car a Travling at v miles per hour is given.

irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

The braking distance at 25 mph, is approximately 63.7 ft.

The braking distance at 55 mph, is approximately 298.35 ft.

The braking distance at 85 mph, is approximately 708.92 ft.

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

$D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.$

At 25 mph, the braking distance is approximately 63.7 ft.

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945 ft.s

$D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.$

At 55 mph, the braking distance is approximately 298.35 ft.

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

$D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.$

At 85 mph, the braking distance is approximately 708.92 ft.

b. The speed of the car when the braking distance is 450 feet is given as follows;

$450 = 2.6 + \dfrac{v^2}{22}$

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ 98.211 ft./s

Learn more here:

brainly.com/question/18591940

8 0

2 years ago

Simplify: (-20)4 ÷ (-20)1 A.(-20)4 B. (-20)5 C. -201 D.(-20)3

Bezzdna [24]

A. (-20)4
Hope this helped! With love, Eilyssa! :)

3 0

3 years ago

Read 2 more answers

A piece of cardboard is 13 inches by 26 inches. A square is to be cut from each corner and the sides folded up to make an open-t

Vanyuwa [196]

Answer:

Hence the maximum possible volume will be the 778.53 c.c

Step-by-step explanation:

Given:

A rectangle with 13 x 26 dimensions

And corners are cut to form side squares.

To Find:

Maximum possible volume for box

Solution :

Consider a rectangle of 13 x 26 dimension with and side of square at corner be x.

(Refer the attachment)

Now,

Formulating the volume equation for the box

So corner square sides we are going to fold up which makes height of the box

and remaining part will be length and breadth

As shown in fig,

Length=26-x

breadth=13-x

And height will be x

$V(x)=x*(26-x)*(13-x)$

To get maximum volume differentiate the above equation,

$V(x)=x*(26*13-26*x-13*x+x^2)$

$V(x)=x^3-39x^2+338x\\$

$V'(x)=3x^2-78x+338$

$V''(x)=6x-78$

Now ,Solve the Quadratic Equation to get x values,

$3x^2-78x+338$ =0

x=[-b±(b^2-4ac)^1/2]/2a

x=[78±Sqrt[(78)^2-4*338*3)]/2*3

x=[78±Sqrt(3028)]/6

x=[78±55.027]/6

x=78+55.027/6 or x=78-55.027/6

x=22.17 or x=3.8288

Use these values in 6x-78 to know which value posses the max and min value for the function.

So when x=22.17

6x-78=6*22.17-78

=55.02>0 i.e function will have minimum value .

When x=3.8288

6*3.8288-78

=-55.0272<0 i.e. Function will have maximum value

Now, the function will defines the maximum volume

$V(x)=x^3-39x^2+338x$

$V(x)=3.8288^3-39*(3.82883)^2+338*3.8288$

$V(x)=56.13-571.73+1294.13$

$V(x)=778.53 C.C$

6 0

3 years ago