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juin [17]
3 years ago
12

Three local gyms are having membership specials.

Mathematics
2 answers:
Snowcat [4.5K]3 years ago
8 0

Answer:

Highest is Able bodies and lowest is, health sytles

Step-by-step explanation:

i took the test.

Reil [10]3 years ago
4 0

Answer:

Lowest: Health Styles

Highest: Able

Able is 32.95$ per month

Health Styles is 20$ per month

Crossfit is 24.95 per month

So, Health is the lowest and Able is the highest.

I figured this out by divided the amount by the amount of months. For example, 359.82 divided by 18 is 19.99 which is also the amount per month.

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Solve: 3n - 6 = -8(6 +5n)<br> and explain how to solve it
hjlf

Answer:

Step-by-step explanation:

First use distributive property: a(b + c) = a*b + a*c

3n - 6 = -8(6 + 5n)

3n - 6 = 6 * (-8) + 5n *(-8)

3n - 6 = -48 - 40n

Now add 40n to both sides

3n - 6 + 40n = -48 - 40n + 40n

43n - 6 = -48

Now add 6 to both sides

43n - 6 + 6 = -48 + 6

           43n = -42

               n = -42 /43

4 0
3 years ago
Read 2 more answers
What type of number is Square root of two
DochEvi [55]

Answer:

irrational

The square root of 2 is irrational.

Step-by-step explanation:

5 0
3 years ago
Line Segments <br> Find ZX
devlian [24]

Answer:

A:4.5

Step-by-step explanation:

all you need to do is count the spaces

4 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
Justin solved this problem.
aleksklad [387]
The answer is A, yes his answer is reasonable because 6 times 0.74 is 44.4
4 0
3 years ago
Read 2 more answers
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