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Svetradugi [14.3K]
3 years ago
14

Write a linear equation from the points (-4,-13) and (3,1)

Mathematics
1 answer:
sveta [45]3 years ago
8 0
Y+13=-12(x+4) is the answer
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Choose whether each decimal terminates or repeats.
Maurinko [17]

Answer:

Step-by-step explanation:

3.232323-repeats

1.789-terminates

0.99-repeats

6.25-terminates

7 0
3 years ago
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Jeffery wants to save the same amount of money each week to buy a new
Temka [501]

Answer:$18

Step-by-step explanation:

7 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
Solve for x ‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎ ‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎ ‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎ ‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎​
inysia [295]

Answer:

12.699

Step-by-step explanation:

i think if you do 26-22.4, you get the leg of the small triangle which is 3.6

then do Pythagorean theorem

3.6^2 + x^2 = 13.2^2

12.96 +x^2 = 174.24

174.24-12.96

x^2 = 161.28

take square root

x = 12.699

6 0
2 years ago
Which equation shows that 30 is 6 times as many as 5?
irina [24]

Answer:

30 divided by 5

Step-by-step explanation:

7 0
3 years ago
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