All categories

3 years ago

Please help me find the slope and y intercept

Mathematics

1 answer:

lbvjy [14]3 years ago

5 0

Answer:

y= -7x +4

Step-by-step explanation:

you find the slope by the slope equation y2-y1/x2-x1. I used (0,4) and (-1,11).

The slope equals -7.

You then use point slope to find the rest of the equation.

y-y1=m(x-x1).

y-4=-7(x-0)

y-4=-7x

y=-7x+4

You might be interested in

5x-2 what’s the slope and y-int ?

uranmaximum [27]

Answer:

Slope = 5.

y-int = (0, -2).

Step-by-step explanation:

y = 5x - 2

The slope intercept form is

y = mx + b where m = the slope and b = the y-intercept.

So here, the slope = 5 and the y-intercept is where y = -2. That is the point (0, -2).

8 0

3 years ago

If the the area of a bedroom is 90 square feet what is the length and width

Anna [14]

30 x 30........................

6 0

3 years ago

A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot

ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = $(x\times x)$

$=x^2$

The volume of the box is = area of the base × height

$=x^2h$

Therefore,

$x^2h=272$

$\Rightarrow h=\frac{272}{x^2}$

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

$=20x^2$ cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

$=10x^2$ cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

$=4xh\times 1.5$ cents

$=6xh$ cents

Total cost = $(20x^2+10x^2+6xh)$

$=30x^2+6xh$

Let

C $=30x^2+6xh$

Putting the value of h

$C=30x^2+6x\times \frac{272}{x^2}$

$\Rightarrow C=30x^2+\frac{1632}{x}$

Differentiating with respect to x

$C'=60x-\frac{1632}{x^2}$

Again differentiating with respect to x

$C''=60+\frac{3264}{x^3}$

Now set C'=0

$60x-\frac{1632}{x^2}=0$

$\Rightarrow 60x=\frac{1632}{x^2}$

$\Rightarrow x^3=\frac{1632}{60}$

$\Rightarrow x\approx 3$

Now $C''|_{x=3}=60+\frac{3264}{3^3}>0$

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is $=\frac{272}{3^2}$

=30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

7 0

3 years ago

Please help pls pls help pls belp

Delvig [45]

Answer:

After 1 second, the ball will reach a maximum height of 16 feet

Step-by-step explanation:

The height of the ball after t seconds: h(t) = -16t^2 + 32t

The graph of this quadratic function is parabola which opens downwards. The vertex of a quadratic equation is the maximum or minimum point on the equation's parabola

t = -b/2a = -(32)/2(-16) = -32/-32 = 1 second

then

h(t) = -16(1)^2 + 32(1) = -16 + 32 = 16

After 1 second, the ball will reach a maximum height of 16 feet

8 0

2 years ago

What is slope of the line

aleksley [76]

3 is the correct slope of the line:)

3 0

2 years ago

Read 2 more answers