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3 years ago

Please help! I don't understand this!! Solve using substitution:

1 answer:

3 0

First you solve for y for the top equation, making the top equation become y=-3+5=x.
now we have to substitute that into 3x-8y=24.

that gives you 3x-8(-3+5x)=24
now you solve it for x, which is 0.
now you have to substitute 0 for x in the equation y=-3+5x, so it now becomes y=-3+5*0.
when you solve this for y, you get -3. this means your answer is y=-3 and x=0

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A trough is 14 ft long and its ends have the shape of isosceles triangles that are 2 ft across at the top and have a height of 1

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4 years ago

PLS NEED HELP! 40 point reward if correct. with brainiest!

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Answer:

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Step-by-step explanation:

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2 years ago

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Angles 1 and 2 form a right angle.Which word describes their measures? linear congruent complementary supplementary

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4 years ago

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Yesterday your bus ride to school took 10 minutes. Today your bus ride took 12 minutes. What is the percent of change?

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The percent of increase from 10 to 12 is 20 %. 10 x 1.2 = 12.

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3 years ago

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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 $121-\left(\dfrac{11}{\sqrt{3}}\right)^2=121-\dfrac{121}{3}=\dfrac{242}{3}\ un.$

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4 years ago

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