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Karo-lina-s [1.5K]
3 years ago
5

Factor-20x2 + 13x + 21. How do you solve this

Mathematics
1 answer:
zaharov [31]3 years ago
5 0

Answer:

-3(9x-7)

Step-by-step explanation:

first multiply

20x2=40  

and now its -40x+13x+21

then add same variables

-40x+13x=-27x

now its -27x+21

then rewrite the equation

3 x 7=21

-3 x 9x = 27x

now its (-3 x 9x) + (3 x 7)

then the multiply thingy and you got the answer

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Solve ln(6x-1) = -4 for x
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x

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A mathematics teacher wanted to see the correlation between test scores and homework. The homework grade (x) and test grade (y)
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2 years ago
PLEASE HELP ASAP! I don’t recall how to do this!
MakcuM [25]

Answer:

Step-by-step explanation:

For a. we start by dividing both sides by 200:

(1.05)^x=1.885

In order to solve for x, we have to get it out from its position of an exponent.  Do that by taking the natural log of both sides:

ln(1.05)^x=ln(1.885)

Applying the power rule for logs lets us now bring down the x in front of the ln:

x * ln(1.05) = ln(1.885)

Now we can divide both sides by ln(1.05) to solve for x:

x=\frac{ln(1.885)}{ln(1.05)}

Do this on your calculator to find that

x = 12.99294297

For b. we will first apply the rule for "undoing" the addition of logs by multipllying:

ln(x*x^2)=5

Simplifying gives you

ln(x^3)=5

Applying the power rule allows us to bring down the 3 in front of the ln:

3 * ln(x) = 5

Now we can divide both sides by 3 to get

ln(x)=\frac{5}{3}

Take the inverse ln by raising each side to e:

e^{ln(x)}=e^{\frac{5}{3}}

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\frac{1}{2}=e^{.1x}

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ln(.5)=ln(e^{.1x})

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(2^2)^x-6(2)^x=-8

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(2^2)^x-6(2)^x+8=0

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2^x=4,2^x=2

For the first solution, we will change the base of 4 into a 2 again like we did in the beginning:

2^2=2^x

Now that the bases are the same, we can say that

x = 2

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2^x=2^1

Now that the bases are the same, we can say that

x = 1

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