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3 years ago

How to solve 36x+3(34x+9)

Mathematics

1 answer:

Softa [21]3 years ago

7 0

Answer: 138x + 27

Step-by-step explanation:

First you distribute the 3, creating 102x +27.

Then you combine like terms, creating 138x + 27

You can't do anything else so that's your answer.

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Which angles are coterminal with −6π5 ? Select each correct answer. 14π/5 −16π/5 −π/5 −11π/5

wariber [46]

ANSWER

$- \frac{16 \pi}{5} \: and \: \frac{14\pi}{5}$

EXPLANATION

Coterminal angles are angles in standard position that have the same terminal side.

To find an angle in standard position that is coterminal with

$- \frac{6 \pi}{5}$

We add or subtract multiples of
$2 \pi$

The first addition gives,

$- \frac{6 \pi}{5} + 2 \pi = \frac{4 \pi}{5}$

We add the second multiple to get,

$- \frac{6 \pi}{5} +2( 2 \pi )= \frac{14 \pi}{5}$

Since this is the maximum value among the options, we end the addition here.

Let us now subtract the first multiple to get,

$- \frac{6 \pi}{5} - 2 \pi = - \frac{16 \pi}{5}$

We end the subtraction here because this is the least value among the options.

Therefore the angles that are coterminal with
$- \frac{6 \pi}{5}$
are

$- \frac{16 \pi}{5} \: and \: \frac{14\pi}{5}$

8 0

3 years ago

Read 2 more answers

How much would Howard Steele need to invest today so that he may withdraw $12,000 each year for the next 20 years, assuming a ra

morpeh [17]

I think its <span>$117,817.20</span>

3 0

3 years ago

(3^2 + 9x + 6) − (8^2 + 3x − 10) + (2x + 4)(3x − 7)

inna [77]

(12+9x) - (6+3x) + 6x + -3
6+3x + 6x+-3
3+ 9x

3 0

3 years ago

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.

8 0

3 years ago

When 2 lines cross, 2 pairs of vertical angles are formed. What is the sum of all 4 angles?

77julia77 [94]

If 2 Lines are crossing together to make 2 pairs of Vertical Angles, the sum of all 4 Angles will always be equal to 360°.

4 0

3 years ago