gibsonj338

$8\frac{3}{8}$

$\frac{67}{8}$

$8.375$

$8.375\times100$

837.5%

$\frac{1+\sqrt{5}}{2}$

$\frac{1}{2}+\frac{\sqrt{5}}{2}$

$0.5+\frac{\sqrt{5}}{2}$

$0.5+\frac{2.2360679774997896964091736687...}{2}$

$0.5+1.1180339887498948482045868343...$

$1.618034...$

$\frac{x}{55}=-11$

$55\times\frac{x}{55}=-11\times55$

$\frac{55x}{55}=-11\times55$

$\frac{1x}{1}=-11\times55$

$\frac{x}{1}=-11\times55$

$x=-11\times55$

$x=-605$

Also, to find out what B is, you cannot take an inverse cosine number in degrees and get an answer in degrees.  You can only take a inverse cosine number in radians and get a number in degrees.  The answer has to be in degrees.

Read my post to you.  You can disagree with me all you want, but you are wrong.  No degree symbol next to the number mean radians,  Degree symbol next to the number means degrees.

EDIT: 0.2345 not 9.2345.

Your answer of 76.437848672801 degrees is incorrect because 9.2345 is not in degrees but in radians.  In order for 0.2345 to be in degrees instead of radians, it need to have the degree symbol: 0.2345°.

$cos(B)=0.2345$

${cos}^{-1}(cos(B))={cos}^{-1}(0.2345)$

$B= 1.334092132482°$

To get the three answers, you need to know what the interest rate formula is.  For this math problem, the interest rate formula is:

$A=P{(1+\frac{r}{n})}^{nt}$

A = amount of money earned after investment

P = amount of money invested

r = interest rate per period

n = number of times the interest is compounded each year

t = number of years the amount of money is deposited for

$A=$10,000{(1+\frac{0.048}{2})}^{2\times3}$

$A=$10,000{(1+\frac{0.048}{2})}^{6}$

$A=$10,000{(1+0.024)}^{6}$

$A=$10,000{(1.024)}^{6}$

$A= $11,529.21504606846976$

$A≈ $11,529.22$

Depositing $10,000 into an account that earns 4.8% for 3 years semiannually wil give you approximately $11,529.22.

$A=$10,000{(1+\frac{0.048}{4})}^{4\times3}$

$A=$10,000{(1+\frac{0.048}{4})}^{12}$

$A=$10,000{(1+0.012)}^{12}$

$A=$10,000{(1.012)}^{12}$

$A=$10,000(1.153894624182586)$

$A=$11,538.94624182586$

$A≈$11,538.95$

Depositing $10,000 into an account that earns 4.8% for 3 years quarterly wil give you approximately $11,538.95.

$A=$10,000{(1+\frac{0.048}{12})}^{12\times3}$

$A=$10,000{(1+\frac{0.048}{12})}^{36}$

$A=$10,000{(1+0.004)}^{36}$

$A=$10,000{(1.004)}^{36}$

$A=$10,000(1.1545524338367405)$

$A= $11,545.524338367405$

$A≈$11,545.52$

Depositing $10,000 into an account that earns 4.8% for 3 years monthlty wil give you approximately $11,545.52.

$({12mn}^{2})({2m}^{2}n)(mn)$

${12mn}^{2}({2m}^{2}n)mn$

$12\times2\times{m}^{2}\times m\times m\times{n}^{2}\times n\times n$

$24\times{m}^{2}\times m\times m\times{n}^{2}\times n\times n$

$24\times{m}^{3}\times m\times{n}^{2}\times n\times n$

$24\times{m}^{4}\times{n}^{2}\times n\times n$

$24\times{m}^{4}\times{n}^{3}\times n$

$24\times{m}^{4}\times{n}^{4}$

$24{m}^{4}{n}^{4}$