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3 years ago

Explain the steps necessary for solving the following equation. x^2-8x-20=0

Mathematics

2 answers:

kirza4 [7]3 years ago

8 0

First factor the equation
(x-10)(x+2) = 0
Then solve for (x-10) and (x+2)
x=10
x=-2

hodyreva [135]3 years ago

3 0

(warning: this is quite long :P)
the steps are:

Looking at the expression x² +8x-20,we can see that the first coefficient is 1, the second coefficient is 8 and the last term is -20.
Now multiply the first coefficient (1) by the last term (-20) to get (1)·(-20)=-20.
Now the question is: what two whole numbers multiply to -20 (the previous product).
factors of -20:
1,2,4,5,10,20
-1, -2, -4, -5, -10, -20
note: list the negative of each factor. This will allow us to find all possible combinations.
These factors pair up to multiply to -20.
1*(-20) = -20
2*(-10) = -20
4*(-5) = -20
(-1)*(20*) = -20
(-2)*(10* = -20
(-4)*(5) = -20
now let's add up each pair of factors to see if one pair adds to the middle coefficient:

first number second number sum

   1    -20 1+(-20)=-19
   2 -10 2+(-10)=-9
   4 -5 4+(-5)= -1
-1 20 -1 +20= 19
-2 10 -2+10=8
-4 5 -4+5=1

From the table above, we can see that the two numbers -2 and 10 multipy to -20 and add up to 8.
Now replace the middle term 8x with -2x+10x. Remember, -2 and 10 add up to 8. so this shows us that -2x+10x=8x.

x²+-2x+10x -20 Replace the second term 8x with -2x +10x.
(x²-2x)+(10x-20) Group the terms into two pairs.
x(x-2x)+(10x-20) Factor out the GCF x from the first group.

x(x-2)+10·(x-2) Factor out 10 from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.

(x+10)·(x-2) Combine like terms. Or factor out thge common term x-2.
-----
ANSWER:

So, x² + 8·x -20 factors to (x+20)·(x-2.
In other words, x² + 8·x-20=(x+10)·(x-2)

THIS IS HARD TO UNDERSTAND, BUT HOPE THIS HELPED YOU! AND HOPE YOU GET IT TOO!!!
:D

Hope it helps :)

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Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

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Zinaida [17]

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What is the arc length when θ = and the radius is 7 cm? (5 points)

maw [93]

What is the arc length when Θ=3 pi/5 and the radius is 7 cm?
Here are the available answers... 21pi/5 cm 12pi/5 cm 6pi/5 cm 3pi/35 cm

Given:
arc length = theta * radius
arc length = (3 pi/5)(7cm)
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4 0

3 years ago

Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$