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Eva8 [605]
1 year ago
7

a utility pole is 8m high. a cable is stretched from the top of the pile to a point in the ground that is 7 m from the bottom of

the pole. how long is the cable? round to 2 decimal places
Mathematics
1 answer:
Olin [163]1 year ago
5 0

We can draw this as:

We can use the Pythagorean theorem to find the length of the cable, as it is the hypotenuse of a right triangle:

\begin{gathered} c^2=7^2+8^2 \\ c^2=49+64 \\ c^2=113 \\ c=\sqrt[]{113} \\ c\approx10.63 \end{gathered}

Answer: the cable length is 10.63 m.

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Can someone help me asap <br>I attempted to do it so just ignore that part ​
Sladkaya [172]

Answer:

a = 180° -73°

b = 73°

c = 180° -73°

Step-by-step explanation:

73° and b are vertical angles, so are congruent (equal).

a and c are equal to each other and each is a linear angle with 73°, so is the supplement of that angle. You could write the equation as ...

a + 73° = 180°

but I prefer to get to the heart of it. The values of a and c are the difference between 180° and 73°.

6 0
3 years ago
A rowing team rowed 90 miles while going with the current in the same amount of time as it took to row 10 miles going against th
lys-0071 [83]

Answer:   
 let x=rate of the rowing team in still water 
 x+4=rate of the rowing team with current 
 x-4=rate of the rowing team against current   
 travel time=distance/rate 
 .. 
 90%2F%28x%2B4%29=10%2F%28x-4%29 
 90x-360=10x+40 
 80x=400 
 x=5  rate of the rowing team in still water=5 mph

4 0
3 years ago
Jerry had 45 tickets for games at a carnival.he used 1/3 of the tickets to play ball-toss game. He used 2/5 of the tickets to pl
MrRa [10]
45 / 3 = 15
45 - 15 = 30
30 / 5 = 6
6 x 2 = 12
30 - 12 = 18
The answer is D:18
7 0
2 years ago
WHAT IS THE SQUARE ROOT OF 400
snow_tiger [21]
The square root of 400 is 20
7 0
3 years ago
Read 2 more answers
1. In an auditorium, there are 21 seats in the first row and 26 seats in the second row. The number of seats in a row continues
kondor19780726 [428]
1. Let s_n be the number of seats in the n-th row. The number seats in the n-th row relative to the number of seats in the (n-1)-th row is given by the recursive rule

s_n=s_{n-1}+5


Since s_1=21, we have

s_2=s_1+5
s_3=s_2+5=s_1+2\cdot5
s_4=s_3+5=s_1+3\cdot5
\cdots
s_n=s_{n-1}+5=\cdots=s_1+(n-1)\cdot5

So the explicit rule for the sequence s_n is

s_n=21+5(n-1)\implies s_n=5n+16

In the 15th row, the number of seats is


s_{15}=5(15)+16=91

2. Let p_n be the amount of profit in the n-th year. If the profits increase by 6% each year, we would have

p_2=p_1+0.06p_1=1.06p_1
p_3=1.06p_2=1.06^2p_1
p_4=1.06p_3=1.06^3p_1
\cdots
p_n=1.06p_{n-1}=\cdots=1.06^{n-1}p_1

with p_1=40,000.

The second part of the question is somewhat vague - are we supposed to find the profits in the 20th year alone? the total profits in the first 20 years? I'll assume the first case, in which we would have a profit of


p_{20}=1.06^{19}\cdot40,000\approx121,024

3. Now let p_n denote the number of pushups done in the n-th week. Since 3\cdot4=12, 12\cdot4=48, and 48\cdot4=192, it looks like we can expect the number of pushups to quadruple per week. So,

p_n=4p_{n-1}

starting with p_1=3.

We can apply the same reason as in (2) to find the explicit rule for the sequence, which you'd find to be

p_n=4^{n-1}p_1\implies p_n=4^{n-1}\cdot3
6 0
3 years ago
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