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Elena-2011 [213]
3 years ago
7

A number x is more than -6 and at most -1

Mathematics
2 answers:
Galina-37 [17]3 years ago
8 0

Answer:

-6<x(less than but equal to sign its not on my keyboard)-1

Step-by-step explanation:

WINSTONCH [101]3 years ago
3 0

Answer:

-6<x≥-1

Step-by-step explanation:

-6<x≥-1

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Graph a line that is perpendicular to the line y=-3x + 2 and has a y-intercept of -6.
Fiesta28 [93]

Answer:

Step-by-step explanation:

the line's equation would be

y = 1/3x -6

so just graph a line using this equation

hope this helps <3

6 0
3 years ago
The value of y varies directly with x. If x = 2, then y = 10. What is the value of x when y = 100? *​
Fynjy0 [20]

Answer:

20

Step-by-step explanation:

y=kx

X=2 & y= 10

10=2k

k=5

when y=100

100=5x

X=20

8 0
3 years ago
Which is not a characteristic of the linear parent function?
GuDViN [60]
B

The range of the parent linear function is All Real Numbers which includes negative numbers
5 0
3 years ago
Write an equation of a line in slope intercept form with the given slope and the y-intercept.
defon
The way you put it is confusing. I'll answer but the way you put it was confusing to me.
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
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