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strojnjashka [21]
3 years ago
8

Solve the two step equation 8x+38=60

Mathematics
1 answer:
Paladinen [302]3 years ago
8 0

Answer:

x = 2.75

Step-by-step explanation:

The goal here is to isolate the variable, or to get x by itself. We can do this by using simple math. The first step is to move every thing to the opposite side of x. We do this by subtracting 38 from 60, which leaves us with 22. Now we want divide 22 by 8. After doing this we are left with the equation x = 2.75. I know that at first fractions and decimals seem annoying and they feel like a lot of work. But if you take a second to really look at them, you will realize they are not that bad. As in most cases with math, looks can be deceiving

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Joelle would like to tip her hairdresser 20%. If her haircut was $38.00, which expression should she use to find the tip amount?
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I pretty sure you would use x over 38.00 = 20 over 100 then cross multiply them
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3 years ago
Solve for p. 2(p + 1) = 16 What is the answer?
Mashutka [201]

Divide both sides by 2

p + 1 = 16/2

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3 years ago
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Find the integral using substitution or a formula.
Nadusha1986 [10]
\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx

Derivative of the denominator:
\rm (x^2+2x+5)'=2x+2

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx

and then split the fraction,

\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx

Based on our previous test, we know that a simple substitution will work for the first integral: \rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx

So the first integral changes,

\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx

integrating to a log,

\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2

So we have this going on,

\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx

Bringing all that stuff together as a single square,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx

Making the substitution: \rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx

giving us,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du

simplying a lil bit,

\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du

and hopefully from this point you recognize your arctangent integral,

\rm ln|x^2+2x+5|+\frac52arctan(u)

undo your substitution as a final step,
and include a constant of integration,

\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c

Hope that helps!
Lemme know if any steps were too confusing.

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3 years ago
HURRY PLSS 50 pt question and brainliest
aksik [14]
The expression is equivalent to 49/8.

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2 years ago
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