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svet-max [94.6K]
3 years ago
5

determine if each pair of rates are equivalent. explain your reasoning $36 for 4 hats; $56 for 7 hats12 posters for 36 students;

21 posters for 63 students
Mathematics
1 answer:
Ugo [173]3 years ago
8 0
36 for 4 hats = 36/4 = 9 per hat
56 for 7 hats = 56/7 = 8 per hat
not equivalent

12 posters for 36 students = 12/36 = 1/3 poster per student
21 posters for 63 students = 21/63 = 1/3 poster per student
these are equivalent...although it does not make much sense...each person getting 1/3 of a poster. And if there was a mistake and it was supposed to be 36 posters for 12 students and 63 posters for 21 students...it would still be equivalent
You might be interested in
Please help me ASAP!!
choli [55]

Answer:

r = -4

Step-by-step explanation:

If M is the midpoint of DE, that would mean that the distance from M to D and M to E would be the same.

This creates the equation 1-8r=13-5r.

Now it's just simple algebra.

You add 8r to both sides, creating 1=13+3r. Then, you subtract 13 from both sides, getting 3r=-12. Dividing both sides by 3 and solving for r, you get r=-4.

Checking our answer, you see that 1-8(-4)=1-(-32)=1+32=33, and 13-5(-4)=13-(-20)=13+20=33.

8 0
3 years ago
Help please thank you!
Maslowich

the error is that they didnt add the exponets

6 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
What is the sum?? need help
SOVA2 [1]
43+35=78 the answer is 78
3 0
3 years ago
What is the numerical coefficient of the a^8*b^2 term in the expansion of ((1/3)a^2 - 3b)^6 ?
zhuklara [117]

In the binomial development, the main problem is calculation of binomial coefficients.

If we want to get term a∧8*b∧2 we see that this is the third member in binomial development (n 2) a∧n-2*b∧2

The given binomial  is ((1/3)a∧2 - 3b)∧6, the first element is (1/3)a∧2, the second element is (-3b) and n=6 when we replace this in the formula we get

(6 2) * ((1/3)a∧2)∧(6-2) * (-3b)2 = (6*5)/2 * ((1/3)a∧2)∧4 *9b∧2= 15*(1/81)*9 *(a∧8b∧2) =

= 15*9* a∧8b∧2 = 135*a∧8b∧2

We finally get numerical coefficient 135

Good luck!!!


7 0
4 years ago
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