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dezoksy [38]
3 years ago
9

The circumference of the inner circle is 66 ft. The distance between the inner circle and the outer circle is 44 ft. By how many

feet is the circumference of outer circle greater than the circumference of the inner​ circle? Use 22/7 for PI.
Mathematics
1 answer:
Ray Of Light [21]3 years ago
6 0

Answer:

  276 4/7 ft

Step-by-step explanation:

Let r1 represent the radius of the inner circle. Then its circumference in feet is ...

  C = 2πr

  66 = 2π(r1)

If the radius of the outer circle is (r1 +44), then its circumference is ...

  C = 2π(r1 +44) = 2π(r1) +2π(44)

  C = 66 +88π

Then the difference in the circumferences is ...

  (66 +88π) - (66) = 88π

Using the given value of π, the difference in circumferences is ...

  (88 ft)(22/7) = 1936/7 ft = 276 4/7 ft

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Louis is saving money for a car, he invests $1,200 in an account that pays an interest rate of 7.5%. How many years will it take
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Answer: n + d = 30

0.05n + 0.10d = 2.20

^^ ur system of equations

Step-by-step explanation:

3 0
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3 SUPER hard math questions HELP!
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Bruce wants to make 50 ml of an alcohol solution with a 12% concentration. He has a 10% alcohol solution and a 15% alcohol solut
Inessa05 [86]

The equation is 0.1x + 0.15(50-x)=1.20(50)

Step-by-step explanation:

Here you need 50ml of 12% alcohol solution.

You have 10% and 15% alcohol solutions.

Take volume of 10% alcohol solution to be x ml

Take volume of 15% alcohol solution to be y ml

Form equations;

x+y=50 ml--------------------------equation 1

0.1x +0.15 y= 12% of 50

0.1x+0.15y=6-----------------------equation 2

Now you have simultaneous equations, use a graph tool to solve for values of x and y as attached.

or use the mixture equation the get value of x

0.1x + 0.15(50-x)=1.20(50)

The values of x and y are ;

(x,y) (30,20) which means;

y is 20 ml of 15% alcohol

x is 30 ml of 10% alcohol

Learn More

  • brainly.com/question/2405496

Keywords : concentration, mixture

#LearnwithBrainly

5 0
3 years ago
If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?
Alisiya [41]
If k is odd, then

\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor

while if k is even, then the sum would be

\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4

The latter case is easier to solve:

\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0

which means k=6.

In the odd case, instead of considering the above equation we can consider the partial sums. If k is odd, then the sum of the even integers between 1 and k would be

S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)

Now consider the partial sum up to the second-to-last term,

S^*=2+4+6+\cdots+(k-5)+(k-3)

Subtracting this from the previous partial sum, we have

S-S^*=k-1

We're given that the sums must add to 2k, which means

S=2k
S^*=2(k-2)

But taking the differences now yields

S-S^*=2k-2(k-2)=4

and there is only one k for which k-1=4; namely, k=5. However, the sum of the even integers between 1 and 5 is 2+4=6, whereas 2k=10\neq6. So there are no solutions to this over the odd integers.
5 0
3 years ago
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