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dezoksy [38]
2 years ago
6

Hey pls help worth 30 points

Mathematics
2 answers:
tino4ka555 [31]2 years ago
8 0

Answer:

y + 8 = 1/2(x + 3)

Step-by-step explanation:

Hope this helpss!!!

Ira Lisetskai [31]2 years ago
3 0

Answer:

i think its 5.0 or -11

Step-by-step explanation:

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Help me plssss hurry!!!!Which expressions are equivalent to 6+(-4) - 5?
Lunna [17]

Answer:

a)

c)

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8 0
2 years ago
Question 1 (5 points)<br>What sign makes the problem true?<br>-15 -16<br>a &lt;<br>b &gt;<br>c =​
Debora [2.8K]
B i cant really tell how because thats hard for me to do but it is b
8 0
3 years ago
-14=-20y-7x<br> 10y+4=2x <br><br> Solve by Elimination please!
iogann1982 [59]

Answer:

x = 2    y = 0

Step-by-step explanation:

8 0
2 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
Im trying to procrastinate so if you would please help me speed up th process. I have like 20 minutes
Katarina [22]

Answer:

1/3=0.33 barred(the threes go on infinitly)

1/6=16666 barred(the sixes go on infinitly)

1/6 as a percent is 16.66% barred(the sixes go on infinitly)

Hope This Helps!!!

7 0
2 years ago
Read 2 more answers
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