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2 years ago

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By how much is the product of 9/5 and 8 1/4 greater than 5

1 answer:

Oliga [24]2 years ago

3 0

Given:

The product of $\dfrac{9}{5}$ and $8\dfrac{1}{4}$ greater than 5.

To find:

By how much the product of $\dfrac{9}{5}$ and $8\dfrac{1}{4}$ greater than 5.

Solution:

First we need to find the product of $\dfrac{9}{5}$ and $8\dfrac{1}{4}$ .

$\text{Product}=\dfrac{9}{5}\times 8\dfrac{1}{4}$

$\text{Product}=\dfrac{9}{5}\times \dfrac{8(4)+1}{4}$

$\text{Product}=\dfrac{9}{5}\times \dfrac{33}{4}$

$\text{Product}=14.85$

Now subtract 5 from the product.

$14.85-5=9.85$

Therefore, the product of $\dfrac{9}{5}$ and $8\dfrac{1}{4}$ is 9.85 greater than 5.

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Makovka662 [10]

Answer:

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Step-by-step explanation:

Since we're only finding the interest for the first three months, it's easy to do it by performing the simple interest formula. But first, we need divide 3 by 12, since we calculate interest using years. 3/12 = 1/4 = 0.25

The standard simple interest calculation is done by multiplying the starting amount, by the interest, by the time, then dividing by 100 to put it into a percentage.

1 month = 1/12 or approximately 0.083 of the year.

Let's say P = 1. For the first month, it will be 1 x 3.6 x 0.083 = 0.2988 / 100

The second month, (1 + 0.002988) * 3.6 * 0.083 = 0.299692814 / 100

The third month, (1.002988 + 0.00299692814) x 3.6 x 0.083 = 0.300588297/100

Given the initial amount be 1, those would be the periodic interest rate during the first three months.

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3 years ago

Solve the following equations.<br> log2(x^2 − 16) − log^2(x − 4) = 1

Alenkasestr [34]

Answer:

$x=\frac{4*(2+e)}{e-2}$

Step-by-step explanation:

Let's rewrite the left side keeping in mind the next propierties:

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Therefore:

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Now, cancel logarithms by taking exp of both sides:

$e^{log(\frac{2*(x^{2} -16)}{(x-4)^{2}})} =e^{1} \\\frac{2*(x^{2} -16)}{(x-4)^{2}}=e$

Multiply both sides by $(x-4)^{2}$ and using distributive propierty:

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Substract $16e-8ex+ex^{2}$ from both sides and factoring:

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Multiply both sides by -1:

$(x-4)*(-8-4e-2x+ex)=0$

Split into two equations:

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Solving for $x-4=0$

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The solutions are:

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This is an absurd because log (x) is undefined for $x\leq 0$

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$log(2*((\frac{4e+8}{e-2})^2-16))-log((\frac{4e+8}{e-2}-4)^2)=1$

Which is correct, therefore the solution is:

$x=\frac{4*(2+e)}{e-2}$

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