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3 years ago

One eighth of Tony's books are mystery books. He has 3 mystery books. How many books does Tony have in all?

Mathematics

2 answers:

Bad White [126]3 years ago

6 0

The easy way is to multiply 1/8 by 3: 1x3=3, 8x3=24, 3/24. Tony has 24 books :)

sergeinik [125]3 years ago

4 0

The question is asking to calculate the book that Tony have in all if one eight of his book are 3 mystery books, base on the questions, and on my further calculation, I would say that the answer would be 24 books. I hope you are satisfied with my answer and feel free to ask for more if you have questions

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Archy [21]

Answer:

64 cubic metres

Step-by-step explanation:

The triangular pyramid volume formula resembles the volume formula for any other pyramid or cone:

V = A * H / 3

where:

A is the area of the pyramid's base

H is the height from the base to the point

In other words: the volume of a triangular pyramid is one-third of the product of the base area and the pyramid's height.

8 x 6 / 2 to find area of base = 24

x 8 = 192

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Hope this makes sense. If you would like, I can l1nk to a very useful website, or I can explain it further.

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3 years ago

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Setler79 [48]

Answer:

Below

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Step-by-step explanation:

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

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

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

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nevsk [136]

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