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Tema [17]
3 years ago
5

George earned $50

Mathematics
1 answer:
abruzzese [7]3 years ago
5 0
Well he earned $50 this week and $25 the following week, so 2/5 of $75 is $30
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A41.0m guy wire is attached to the top of a 32.8 m antenna and to a point on the ground. How far is the point on the ground from
Setler79 [48]

Answer:

Step-by-step explanation:

A right angle triangle is formed.

The length of the guy wire represents the hypotenuse of the right angle triangle.

The height of the antenna represents the opposite side of the right angle triangle.

The distance, h from base of the antenna to the point on the ground to which the antenna is attached represents the adjacent side of the triangle.

To determine h, we would apply Pythagoras theorem which is expressed as

Hypotenuse² = opposite side² + adjacent side²

Therefore,

41² = 32.8² + h²

1681 = 1075.84 + h²

h² = 1681 - 1075.84 = 605.16

h = √605.16

h = 24.6 m

To determine the angle θ that the wire makes with the ground, we would apply the the cosine trigonometric ratio.

Cos θ = adjacent side/hypotenuse. Therefore,

Cos θ = 24.6/41 = 0.6

θ = Cos^-1(0.6)

θ = 53.1°

7 0
4 years ago
You are multiplying as many as you want 5s and as many as you want 2s, at least one of each. What could the last digit of the pr
Aleksandr [31]

Answer:

0

Step-by-step explanation:

It is always going to be 0 because there is always going to be a 2 and a 5 that multiply to make the units digit zero.

I hope this could help you.

5 0
3 years ago
Read 2 more answers
From a point on a cliff 75 feet above water level, an observer can see a shipThe angle of depression to the ship is d^ How far i
Alexandra [31]

Answer:

52feet

Step-by-step explanation:

The set up will be a right angle

Let the angle of depression be 30°

The height above the sea level will be the opposite

The distance of the ship from the base is x (adjacent)

Using TOA

Tan theta = opposite/adjacent

Tan 30 = 30/x

x = 30/tan 30

x = 30/0.5774

x = 51.96

x ≈ 52 feet to the nearest foot

8 0
2 years ago
How to factor 4(x+5)^3(x-1)^2-(x+5)^4 • 2 (x-1) by grouping?
nata0808 [166]
\bf 4(x+5)^3(x-1)^2-(x+5)^4\cdot 2(x-1)
\\\\\\
4(x+5)^3(x-1)^2-2(x-1)(x+5)^4
\\\\\\ \underline{2}\cdot 2\underline{(x+5)^3}(x-1)\underline{(x-1)}~-~\underline{2}\underline{(x-1)}(x+5)\underline{(x+5)^3}\leftarrow 
\begin{array}{llll}
notice~the\\
common\\ \underline{factors}
\end{array}
\\\\\\
2(x+5)^3(x-1)~[~2(x-1)-(x+5)~]
\\\\\\
2(x+5)^3(x-1)~[2x-2-x-5]\implies 2(x+5)^3(x-1)(x-7)
5 0
3 years ago
<img src="https://tex.z-dn.net/?f=-5%283.15-%5Cfrac%7B21%7D%7B20%7D%20%29%2B7" id="TexFormula1" title="-5(3.15-\frac{21}{20} )+7
Maksim231197 [3]
The answer is -7/2 or -3.5
8 0
3 years ago
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