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3 years ago

If mZQOS = 49, mZPOR = 65, and mZPOQ = 28, what is mZROS?

Mathematics

1 answer:

dangina [55]3 years ago

6 0

011928221992838220928282

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Which expression is equivalent to the square root of 2^5/18

blsea [12.9K]

1.77777777778 would be the answer you are looking for I think.

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3 years ago

Which point is 10 units from (2,5)

zalisa [80]

Answer:

All the points on (x - 2)² + (y - 5)² = 100

Step-by-step explanation:

It should be the circle with its center on point (2,5) and radius is 10.

(h , k) : center of circle (2,5) h = 2 k = 5

r: radius r = 10

Circle: (x - h)² + (y - k)² = r²

(x - 2)² + (y - 5)² = 100

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3 years ago

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Find BC if B(8,-7) and C1-4,-2).

Alika [10]

Using distance formula :

$\sqrt{(x_2 - x_1) {}^{2} + (y_2 - y_1) {}^{2} }$

$\sqrt{(8 - ( - 4)) {}^{2} + ( - 7 - ( - 2)) {}^{2} }$

$\sqrt{(12) {}^{2} + ( - 5) {}^{2} }$

$\sqrt{144 + 25}$

$\sqrt{169}$

$13 \: \: units$

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3 years ago

A business was valued at £80000 at the start of 2013. In 5 years the value of this business raised to £95000. this is equivalent

Yuri [45]

the yearly increase of x% assumes is compounding yearly, so let's use that.

$~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill &\£95000\\ P=\textit{original amount deposited}\dotfill &\£80000\\ r=rate\to r\%\to \frac{r}{100}\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &5 \end{cases}$

$95000=80000\left(1+\frac{~~ \frac{r}{100}~~}{1}\right)^{1\cdot 5}\implies \cfrac{95000}{80000}=\left( 1+\cfrac{r}{100} \right)^5 \\\\\\ \cfrac{19}{16}=\left( 1+\cfrac{r}{100} \right)^5\implies \sqrt[5]{\cfrac{19}{16}}=1+\cfrac{r}{100}\implies \sqrt[5]{\cfrac{19}{16}}=\cfrac{100+r}{100} \\\\\\ 100\sqrt[5]{\cfrac{19}{16}}=100+r\implies 100\sqrt[5]{\cfrac{19}{16}}-100=r\implies 3.5\approx r$

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2 years ago

Question 2 (1 point)<br> Which graph shows the solutions to 2x<6?

Studentka2010 [4]

Answer: C
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3 years ago

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