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photoshop1234 [79]
3 years ago
14

PLEASE HELP. i’ll give brainliest if you want!

Mathematics
1 answer:
ira [324]3 years ago
3 0

Answer:

sharing of simple moral stories

Step-by-step explanation:

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If two lines are the same slope and diffrent y-intercepts then the lines are
iVinArrow [24]

Answer:

The lines would be parallel

Step-by-step explanation:

They have the same slope, but they sit on different y values. Therefore, they are parallel.

7 0
3 years ago
If the AREA of the rectangle at the right is 22 3/4 square inches, what is the width of the rectangle?
sergejj [24]

Answer:

3.25 or 3 1/4 inches

Step-by-step explanation:

A = lw

22.75 = 7w

w = 3.25 which is 3 1/4

6 0
3 years ago
Find square root of <br><img src="https://tex.z-dn.net/?f=244" id="TexFormula1" title="244" alt="244" align="absmiddle" class="l
cestrela7 [59]

Answer:

15.6204994

Step-by-step explanation:

15.6204994^{2}  = 244\\so\\\sqrt{244} = 15.6204994

hope this helps!

6 0
2 years ago
If a car travels 60 miles at 30 miles per hour and then 60 miles an hour, what was the car's average speed for the total trip?
docker41 [41]
The answer is 45 because if you add 30 and 60 which equals 90 then you do 90 divided by 2.
6 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.

The dimensions of the rectangle are:

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

6 0
4 years ago
Read 2 more answers
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