Antonio used 16 1/3 cups of flour to make banana bread. The recipe for one loaf of banana bread calls for 2 1/3

What is the slope of the line that passes through the points (-10, -8) and

RoseWind [281]

Answer:

Equation: y = -4x + 32

Slope: -4

Step-by-step explanation:

Given : (-10,-8) and (-8, -16)

Slope Formula: $\frac{y_1 - y_2}{x_1-x_2}$

Solve For Slope, Input Given Points:

$\frac{-8 - (-16)}{-10 - (-8)}$
$\frac{8}{-2}$
$-4$

Solve for y-intercept:

Input the slope into the equation y = mx + b. Plugin x and y values for the x and y variables to find b.

y = -4x + b
-8 = -10(4) + b
-8 = -40 + b
b = 32

Equation of LIne: y = -4x + 32

-Chetan K

5 0

2 years ago

Read 2 more answers

Four friends go out to eat and their meal costs $86.50. The sales tax is 6.75% and they should tip the waiter 20%. How much shou

GenaCL600 [577]

If the tax is placed on the roigial price, they should each pay around 49 dollars and 3 cents.

8 0

3 years ago

Read 2 more answers

Is x2y2-4x3+12y a polynomial

masya89 [10]

I'm pretty sure it is.

If i'm wrong, I'm truly sorry

6 0

4 years ago

The line is translated 2 units down which equation will best describes the new line

ss7ja [257]

Answer:

The y intercept should be 2 lower than the original equation

Step-by-step explanation:

6 0

3 years ago

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by

11111nata11111 [884]

Answer:

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is $V=l\times b\times h$

$V=(11-2x)\times (7-2x)\times x$

$V=4x^3-36x^2+77x$

Derivate w.r.t x,

$V'(x)=4(3x^2)-2(36x)+77$

$V'(x)=12x^2-72x+77$

The critical point when V'(x)=0

$12x^2-72x+77=0$

Solve by quadratic formula,

$x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}$

$x=4.607,1.392$

Derivate again w.r.t x,

$V''(x)=24x-72$

Now, $V''(4.607)=24(4.607)-72=38.568>0$ (+ve)

$V''(1.392)=24(1.392)-72=-38.592$ (-ve)

So, there is maximum at x=1.392.

The length of the box is $l=11-2x$

$l=11-2(1.392)=8.216$

The breadth of the box is $b=7-2x$

$b=7-2(1.392)=4.216$

The height of the box is h=1.392.

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is $V=l\times b\times h$

$V=8.216\times 4.216\times 1.392$

$V=48.217\ in.^3$

The volume of the box is 8.217 cubic inches.

5 0

3 years ago