CoolStuffYT

When it falls the first time, it goes 16 meters.

When it goes up the first time, it goes up 16*3/4 = 12 meters.

When it falls the second time, it goes 12 meters.

When if goes up the second time, it goes up 12*3/4 meters = 9 meters.

When it falls the third time, it goes 9 meters.

When it goes up the third time, it goes 9*3/4 = 27/4 meters.

16+12+12+9+9+27/4 = 16+24+18+27/4 = (64+96+72+27)/4 = (160+99)/4 = 259/4.

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Let's call the width "x."

The width is x and the length is 3x - 1

The area of the width and length is 30.

(x)(3x-1) = 30

3x^2-x = 30

3x^2-x-30 = 0

(3x+3)(x-10) = 0

Solve 3x+3=0 and x-10=0 and see which value of "x" is feasible. (hint: widths cannot be negative.)

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If their solution is in common, we can set the equations equal to one another

So 2x^2-13x+21 = 2x^2+9x-56

-22x = -77

Solve x from here and round to the nearest tenth.

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I assume you mean \(3x^2-10x-8\)

You can factor it out as (3x+2)(x-4) = 0

From there you can find 3x+2 = 0 and x-4 = 0, solve for x

The positive x value is your solution.

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Area of the triangle = b*h/2 = (2x+4)(x+3)/2 = 2x^2+10x+12 all divided by 2 = x^2+5x+6.

\(x^2+5x+6 = 56, x^2+5x-50 = 0, (x-5)(x+10) = 0\)

can you solve from here?

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First thing is to find the roots.

We can use the quadratic formula to find the roots

\(\frac{-21\pm\sqrt{21^2-4(18)(-400)}}{2\times 18} = \frac{-21\pm 171}{36}\)

x equals 25/6 or -16/3.

\((\frac{25}{6})^2+(-\frac{16}{3})^2=\frac{625}{36}+\frac{256}{9}=\frac{1649}{36}\)

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f(g(7))+g(f(3)). Seems complex at first.

However, we can evaluate g(7), which is 7+7 = 14, and now we can find f(14)

Also, we can evaluate f(3), which is 3^2-1=8, and now we can find g(8).

f(14)=14^2-1=195 and g(8)=8+7=15.

Now we have 195+15 which equals 210.

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Thanks Chris!

If 3/20 trees are infested and there are 1200 trees total, then 3/20*1200=180 so there should be around 180 infested trees.

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Let's see.

(x+1)(x+1)-(x+2)(x-3)

(x^2+2x+1)-(x^2-x-6)

x^2+2x+1-x^2+x+6

3x+7

We can see how he didn't distribute the negative sign in -(x^2-x-6) to the -x and -6.

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