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Ivenika [448]
3 years ago
6

How much of an alloy that is 10% copper should be mixed with 600 ounce of an alloy that is 70%copper in order to get an alloy th

at is 30%copper?
Mathematics
1 answer:
svetoff [14.1K]3 years ago
7 0

Answer:

1200 ounces

Step-by-step explanation:

How much of an alloy that is 10% copper should be mixed with 600 ounce of an alloy that is 70%copper in order to get an alloy that is 30%copper?

Let us represent

Number of ounce of 10% copper = x

Hence, our Equation =

x × 10% + 600 × 70% = (x + 600) × 30%

0.1x + 420 = (x + 600)0.3

0.1x + 420 = 0.3x + 180

Collect like terms

0.3x - 0.1x = 420 - 180

0.2x = 240

x = 240/0.2

x = 1200 ounces

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First, we have to see that this is an arithmetic sequence... since to get the next element we add 5 to it.  (a geometric sequence would be a multiplication, not an addition)

So, we have a, the first term (a = 4), and we have the difference between each term (d = 5), and we want to find the SUM of the first 100 terms.

To do this without spending hours writing them down, we can use this formula:

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If we plug in our values, we have:

S = \frac{100}{2} * (2 * 4 + (100 - 1) * 5) = 50 * (8 + 99 * 5)

S = 50 * (8 + 495) = 50 * 503 = 25150

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If you're using the app, try seeing this answer through your browser:  brainly.com/question/3151018

——————————
 
     \mathsf{\dfrac{3\sqrt{6}+5\sqrt{2}}{4\sqrt{6}-3\sqrt{2}}}


Multiply and divide by the conjugate of the denominator, which is  \mathsf{4\sqrt{6}+3\sqrt{2}:}

     \mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}}


Expand those products and eliminate the brackets:
 
     \mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(3\sqrt{6}+5\sqrt{2}\big)\cdot 3\sqrt{2}}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(4\sqrt{6}-3\sqrt{2}\big)\cdot 3\sqrt{2}}}\\\\\\ \mathsf{=\dfrac{12\cdot \big(\sqrt{6}\big)^2+20\sqrt{2}\cdot \sqrt{6}+9\cdot \sqrt{6}\cdot \sqrt{2}+15\cdot \big(\sqrt{2}\big)^2}{16\cdot \big(\sqrt{6}\big)^2-12\cdot \sqrt{2}\cdot \sqrt{6}+ 12\cdot \sqrt{6}\cdot \sqrt{2}-9\cdot \big(\sqrt{2}\big)^2}}\\\\\\ \mathsf{=\dfrac{12\cdot 6+20\sqrt{2\cdot 6}+9\sqrt{6\cdot 2}+15\cdot 2}{16\cdot 6-12\sqrt{2\cdot 6}+ 12\sqrt{6\cdot 2}-9\cdot 2}}\\\\\\ \mathsf{=\dfrac{72+20\sqrt{12}+9\sqrt{12}+30}{96-12\sqrt{12}+12\sqrt{12}-18}}

     \mathsf{=\dfrac{72+29\sqrt{12}+30}{96-18}}\\\\\\ \mathsf{=\dfrac{102+29\sqrt{2^2\cdot 3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot \sqrt{2^2}\cdot \sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot 2\sqrt{3}}{78}} 

     \mathsf{=\dfrac{102+58\sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{\,\diagup\!\!\!\! 2\cdot \big(51+29\sqrt{3}\big)}{\diagup\!\!\!\! 2\cdot 39}} 

     \mathsf{=\dfrac{51+29\sqrt{3}}{39}}\quad\longleftarrow\quad\textsf{this is the answer.}


I hope this helps. =)

4 0
3 years ago
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