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sdas [7]
3 years ago
7

How much of a 29% solution is required to mix with 44mL of a 51% solution to make a 37% solution?

Mathematics
1 answer:
nirvana33 [79]3 years ago
4 0
Let me take a stab at this...these are quite confusing and difficult! I made a table like this:
           mL sol.                %                  amt%
29%      x                      .29      =        .29x
51%      44                    .51      =      22.44
--------------------------------------------------------
37%    44+x                  .37      =  .37(44+x)
You have to add the first two rows together to give you what you want in the end. In other words, how much (x) of a 29% solution (.29) + 44mL of 51% (.51) solution will give you a solution of 37% (.37). The word "of" in those phrases mean to multiply:  .29x + 22.44 = .37(44+x). Distribute the .37 into the parenthesis to get .29x + 22.44 = 16.28 + .37x. Solve for x to get x=77 mL
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What is the probability of getting blue and green on a spinner that is 3/10 and 1/5 blue
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5 0
3 years ago
5^(-x)+7=2x+4 This was on plato
Setler79 [48]

Answer:

Below

I hope its not too complicated

x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

Step-by-step explanation:

5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}

Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}

\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x

\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

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mrs_skeptik [129]

Answer:steepness

Step-by-step explanation:

5 0
3 years ago
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-6+p=-22

add 6 to each side

+6 -6+p=-22+6

p = -16

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