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3 years ago

What is the better deal: Babysitting for 8 hours and making $72.00 or babysitting for 6 hours and making $57.00

1 answer:

5 0

The better deal would be to babysit for 6 hours and making $57.00.
1. First, I found the unit rate. How much do you make for 1 hour of work? 72/8=9. So, you make $9.00 per hour.
2. Second, I did the same thing with the other situation. 57/6= 9.50. So, you make $9.50 per hour.
3. The better deal would be to babysit for 6 hours making $57.00 because you are making more money per hour. 9<9.5

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What is the area of this triangle ?

oee [108]

Answer:

Area of triangle is 9.88 units^2

Step-by-step explanation:

We need to find the area of triangle

Given E(5,1), F(0,4), D(0,8)

We will use formula:

$Area\,\,of\,\,triangle =\sqrt{s(s-a)(s-b)s-c)} \\where\,\, s = \frac{a+b+c}{2}$

We need to find the lengths of side DE, EF and FD

Length of side DE = a = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Length of side DE = a = $=\sqrt{(5-0)^2+(1-8)^2}\\=\sqrt{(5)^2+(-7)^2}\\=\sqrt{25+49}\\=\sqrt{74}\\=8.60$

Length of side EF = b = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Length of side EF = b = $=\sqrt{(0-5)^2+(4-1)^2}\\=\sqrt{(-5)^2+(3)^2}\\=\sqrt{25+9}\\=\sqrt{34}\\=5.8$

Length of side FD = c = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Length of side FD = c = $=\sqrt{(0-0)^2+(8-4)^2}\\=\sqrt{(0)^2+(4)^2}\\=\sqrt{0+16}\\=\sqrt{16}\\=4$

so, a= 8.60, b= 5.8 and c = 4

s = a+b+c/2

s= 8.6+5.8+4/2

s= 9.2

Area of triangle= $=\sqrt{s(s-a)(s-b)s-c)}\\=\sqrt{9.2(9.2-8.6)(9.2-5.8)(9.2-4)}\\=\sqrt{9.2(0.6)(3.4)(5.2)}\\=\sqrt{97.5936}\\=9.88$

So, area of triangle is 9.88 units^2

4 0

3 years ago

In the triangle shown we can find the angle θ as follows calculator

Xelga [282]

Given a right triangle with hypothenus of measure 34, the side opposite the angle θ of measure 30, and the side adjacent the angle theta of measure 16.

$\sin\theta= \frac{opposite}{hypothenuse} = \frac{30}{34} = \frac{15}{17} \\ \\ \Rightarrow \theta=\sin^{-1}\left( \frac{15}{17} \right) \\ \\ \\ \cos\theta= \frac{adjacent}{hypothenuse} = \frac{16}{34} = \frac{8}{17} \\ \\ \Rightarrow \theta=\cos^{-1}\left( \frac{8}{17} \right) \\ \\ \\ \tan\theta= \frac{opposite}{adjacent} = \frac{30}{16} = \frac{15}{8} \\ \\ \Rightarrow \theta=\cos^{-1}\left( \frac{15}{8} \right)$

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