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aliina [53]
2 years ago
11

Guys, I beg you! How do you do these factoring quadratic equations!

Mathematics
2 answers:
Romashka-Z-Leto [24]2 years ago
8 0
<h3>Answers:  x = -11 and x = 3</h3>

======================================================

Explanation:

There are at least three ways to solve this.

----------------------------------------------

Method 1:

We could graph y = (x-2)(x+10) using a tool like Desmos. Graph y = 13 as well which is a horizontal line.

Check out the screenshot shown below.

The parabola in red and the horizontal line in blue intersect at the locations (-11, 13) and (3, 13)

We only focus on the x coordinates of the intersection points.

Therefore, the two solutions are x = -11 and x = 3

The order of the solutions doesn't matter.

----------------------------------------------

Method 2:

Expand things out and get everything to one side

(x-2)(x+10) = 13

x^2 + 10x - 2x - 20 = 13

x^2 + 8x - 20 - 13 = 0

x^2 + 8x - 33 = 0

From here we can use the quadratic formula

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-8\pm\sqrt{(8)^2-4(1)(-33)}}{2(1)}\\\\x = \frac{-8\pm\sqrt{64+132}}{2}\\\\x = \frac{-8\pm\sqrt{196}}{2}\\\\x = \frac{-8\pm14}{2}\\\\x = \frac{-8+14}{2} \ \text{ or } \ x = \frac{-8-14}{2}\\\\x = \frac{6}{2} \ \text{ or } \ x = \frac{-22}{2}\\\\x = 3 \ \text{ or } \ x = -11\\\\

----------------------------------------------

Method 3:

Instead of using the quadratic formula, we could factor.

To factor x^2+8x-33, we need to find two numbers that

  • Multiply to -33
  • Add to 8

Through trial and error, you should get 11 and -3

  • 11 times -3 = -33
  • 11 plus -3 = 8

So,

x^2+8x-33 = 0

(x+11)(x-3) = 0

x+11 = 0 or x-3 = 0

x = -11 or x = 3

Valentin [98]2 years ago
5 0
Times the x’a together and x*10 then 2*x and 2*10
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Answer:

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Step-by-step explanation:

5 0
3 years ago
A vertical right circular cylindrical tank has height h=8 feet high and diameter d=6 feet. It is full of kerosene weighing 50 po
Mariana [72]

Answer:

The work done to  pump all of the kerosene from the tank to an outlet is W=45238.9\: J  

Step-by-step explanation:

The work is defined by:

W=\int dFdx (1)    

The force here will be the product between the volume and the kerosene weighing, so we have :

dF=\pi R^{2}dy*50

This force will be in-lbs.

Where R is the radius (3 feet)                    

Then using (1), we have:

W=\int \pi R^{2}dy*50(8-y)  

Here 8-y is a distance at some point of the tank. Now, to get the work done from the base to the top of the tank we will need to take integral from 0 to 8 feet.

W=\int_{0}^{8} \pi 3^{2}dy*50(8-y)

W=450\pi \int_{0}^{8}dy(8-y)

W=450\pi(\int_{0}^{8} 8dy-\int_{0}^{8} ydy)

W=450\pi(8y|_{0}^{8} -\frac{y^{2}}{2}|_{0}^{8})  

W=450\pi(8*8 -\frac{8^{2}}{2})

W=450\pi(64 -\frac{64}{2})

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I hope it helps you!  

6 0
3 years ago
Rewrite the following polynomial in standard form.<br> -x^2-10+x^3
ELEN [110]

Answer:

37

Step-by-step explanation:

4 0
3 years ago
Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.
adoni [48]

Answer: The required derivative is \dfrac{8x^2+18x+9}{x(2x+3)^2}

Step-by-step explanation:

Since we have given that

y=\ln[x(2x+3)^2]

Differentiating log function w.r.t. x, we get that

\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [x'(2x+3)^2+(2x+3)^2'x]\\\\\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [(2x+3)^2+2x(2x+3)]\\\\\dfrac{dy}{dx}=\dfrac{4x^2+9+12x+4x^2+6x}{x(2x+3)^2}\\\\\dfrac{dy}{dx}=\dfrac{8x^2+18x+9}{x(2x+3)^2}

Hence, the required derivative is \dfrac{8x^2+18x+9}{x(2x+3)^2}

3 0
3 years ago
Please help me with these. <br> Thanks
DIA [1.3K]
5.

f(K) = D^3 => f(25) = 125 => 25 * t = 125 ( because K is directly proportional with D^3 )=> t = 125 / 25 => t = 5 => f(25) = 25 * 5 => K * 5 = D^3 ;

6. 

f(L) = F^3 => f(2) =3^3 =>f(2) = 27 => 2 / t =27 => t = 2 / 27 => t = 0.074 => f(2) = 2 / 0.074 => K / 0.074 = F^3 ;


6 0
4 years ago
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