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3 years ago

A linear relationship has the point (4, 48) and its rate of change is 12.

Mathematics

1 answer:

Triss [41]3 years ago

5 0

Answer:

Step-by-step explanation:

Y-48=12(x-4)

Y= 12x

So the y intercept is 0

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Use conversion factors to convert . Write the factor you used . 5 quarts to cups

34kurt

A quart is 4 cups. 5(4)=20 cups in a pint

I hope I helped :)

4 0

3 years ago

Find the slope of the line that passes through the pair of points.<br> (6, 1), (-6, 1)

GrogVix [38]

Answer: m = 0

Step-by-step explanation: To find the slope of the line, we can use the graphing method. To do this, we start by plotting our two points.

Let's label (6, 1) as point A and (-6, 1) as point B.

Now, graph our line through the 2 points.

Now remember that the slope, or m, is equal to

the rise over run from point A to point B.

To get from point A to point B, we rise o units and run -12 units.

So our slope or rise / run is 0/-12 which simplifies to 0.

This should make sense because we have a horizontal or flat line and

remember that the slope of any horizontal or flat line is 0.

Image attached.

7 0

2 years ago

Read 2 more answers

6 students want lemonade and 12 students

Ksenya-84 [330]

Answer:

2:1

Step-by-step explanation:

well we show ratio by the sign ' : '

ratio of iced tea to lemonade= 12:6

now we reduce it to minimum by dividing it with same number.

12 is divisible by 6

so,

the reduced form is = 2:1

therefore,

ratio of the number of students who want iced tea to the number of students who want lemonade = 2:1.

5 0

3 years ago

a rectangle is constructed with its base on the x-axis and two of its vertices on the parabola y=121-x^2. what are the dimension

insens350 [35]

Answer:

the length is $\dfrac{22}{\sqrt{3}}$ and the width is $121-\left(\dfrac{11}{\sqrt{3}}\right)^2=121-\dfrac{121}{3}=\dfrac{242}{3}.$

Step-by-step explanation:

Let points A and B be placed on the x-axis. Their coordinates are $A(x_0,0)\ (x_0>0)$ and $B(-x_0,0)$ (because of parabola symmetry). Two other vertices lie on the parabola, then $C(-x_0,121-x_0^2)$ and $D(x_0,121-x_0^2).$ The length of the side AB is $2x_0$ and the length of the side AD is $121-x_0^2.$ Thus, the area of the rectangle ABCD is

$A=2x_0\cdot (121-x_0^2)=242x_0-2x_0^3.$

Find the derivative A':

$A'=242-2\cdot 3x_0^2=242-6x_0^2.$

Equate A' to 0:

$242-6x_0^2=0,\\ \\x_0^2=\dfrac{121}{3},\\ \\x_0=\dfrac{11}{\sqrt{3}}.$

The maximum area of the rectangle is

$A_{max}=242\cdot \dfrac{11}{\sqrt{3}}-2\cdot \left(\dfrac{11}{\sqrt{3}}\right)^3=\dfrac{2662}{\sqrt{3}}-\dfrac{2662}{3\sqrt{3}}=\dfrac{5324}{3\sqrt{3}}\ un^2.$