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Oliga [24]
2 years ago
9

It takes Dimitri 9 minutes to make a simple bracelet and 20 minutes to make a deluxe bracelet. He has been making bracelets for

longer than 120 minutes. If x represents the number of simple bracelets that he has made and yrepresents the number of deluxe bracelets he has made, the inequality 9x + 20y > 120 represents the scenario. Which is a possible combination of bracelets that Dimitri may have made?
Mathematics
2 answers:
miskamm [114]2 years ago
5 0

Answer:

(1, 14); (2, 9); (3, 7); ...

Step-by-step explanation:

Starting from the relationship between the variables x and y defined, you can obtain:

x> \frac{(120 - 20y) }{9} as long as y is an integer value greater than 0 and less than 6. Therefore, we have infinite combinations that meet the conditions of the statement. Examples of possible combinations for the couple (x, y) are:

(1, 14); (2, 9); (3, 7); ...

maks197457 [2]2 years ago
3 0

3 simple bracelets and 4 deluxe bracelets

0 simple bracelets and 6 deluxe bracelets

12 simple bracelets and 0 deluxe bracelets

7 simple bracelets and 3 deluxe bracelets

Given the inequality 9x + 20y > 120, it is easier to just substituting the answers from the choices and see which ones satisfy the inequality. From the choices, the correct answer is 7 simple bracelets and 3 deluxe bracelets, which yield an answer of 123, which is greater than 120.

Hope this helps!

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Read 2 more answers
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
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