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kirza4 [7]
3 years ago
11

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}

3 0
3 years ago
How to divide 1/8 ÷ 4
Zinaida [17]
Multiply 4 by its reciprocal: 1/8x1/4= 1/32.
8 0
3 years ago
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What is the arc length when θ = and the radius is 7 cm? (5 points)
maw [93]
<span>What is the arc length when Θ=3 pi/5 and the radius is 7 cm?
</span><span>Here are the available answers... 21pi/5 cm 12pi/5 cm 6pi/5 cm 3pi/35 cm
</span>
Given:
arc length = theta * radius
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arc length = 21pi/5 cm   Answer is the 1st option.
4 0
3 years ago
Let z=3+i, <br>then find<br> a. Z²<br>b. |Z| <br>c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq
zysi [14]

Given <em>z</em> = 3 + <em>i</em>, right away we can find

(a) square

<em>z</em> ² = (3 + <em>i </em>)² = 3² + 6<em>i</em> + <em>i</em> ² = 9 + 6<em>i</em> - 1 = 8 + 6<em>i</em>

(b) modulus

|<em>z</em>| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(<em>z</em>) = arctan(1/3)

Then

<em>z</em> = |<em>z</em>| exp(<em>i</em> arg(<em>z</em>))

<em>z</em> = √10 exp(<em>i</em> arctan(1/3))

or

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

(c) square root

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

√<em>z</em> = √(√10) exp(<em>i</em> arctan(1/3) / 2)

and

√<em>z</em> = √(√10) exp(<em>i</em> (arctan(1/3) + 2<em>π</em>) / 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 <em>z</em> lies in the first quadrant, so

0 < arg(<em>z</em>) = arctan(1/3) < <em>π</em>/2

which means

0 < 1/2 arctan(1/3) < <em>π</em>/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(<em>x</em> + <em>π</em>) = -cos(<em>x</em>) and sin(<em>x</em> + <em>π</em>) = -sin(<em>x</em>),

\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 <em>y</em> such that tan(<em>y</em>) = 1/3. In a right triangle satisfying this relation, we would see that cos(<em>y</em>) = 3/√10 and sin(<em>y</em>) = 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}}

\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}

\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}

\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}

So the two square roots of <em>z</em> are

\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}

and

\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}

3 0
3 years ago
Read 2 more answers
HELP ME PLZ!!!! if the answer is right ill mark you brainliest
STALIN [3.7K]

Answer:

9.6558916e+17

Scientific Notation: 3.45 x 10^5

E Notation: 3.45e5

= 9.26 × 1010

(scientific notation)

= 9.26e10

(scientific e notation)

= 92.6 × 109

(engineering notation)

(billion; prefix giga- (G))

its probably: 9.26 × 1010 as the question asks for scientific notation

Step-by-step explanation:

somewhere between those lines, good luck!

5 0
2 years ago
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