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Leviafan [203]
3 years ago
12

What is 1 12/20 as a percentage

Mathematics
2 answers:
agasfer [191]3 years ago
7 0

Step-by-step explanation:

step 1. 1-12/20 = 1.6

step 2. 1.6 = 160%.

Vedmedyk [2.9K]3 years ago
5 0
Its 160 okk like u use photomath
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A kite is 85 feet high with 100 feet of string let out. What is the angle of elevation of the string with the ground? Please sho
a_sh-v [17]

Answer:

  58°

Step-by-step explanation:

A right triangle can be drawn to model the geometry of the problem. The hypotenuse of the triangle is the length of the string, 100 ft. The side opposite the angle is the height of the kite above the ground, 85 ft.

The mnemonic SOH CAH TOA reminds you of the relationship between sides and angles.

  Sin = Opposite/Hypotenuse

  sin(α) = (85 ft)/(100 ft) = 0.85

The angle whose sine is 0.85 is found using the arcsine (inverse sine) function:

  α = arcsin(0.85) ≈ 58.2°

The angle of elevation is about 58°.

_____

When using your calculator to find the values of inverse trig functions, make sure it is in <em>degrees</em> mode. Otherwise, you're likely to get the answer in radians (≈ 1.01599 radians).

6 0
3 years ago
6. Write the equation that represents the points in
luda_lava [24]

Answer:

0. | 2

Step-by-step explanation:

I took (0.2) and made an equal sign and gave me 0. | 2

3 0
3 years ago
2/4 plus 1/3 as a fraction
QveST [7]

Answer: 5/6

Step-by-step explanation: trust me on this

7 0
3 years ago
Read 2 more answers
[100+(586+40/10)]/5&lt;19
Ksivusya [100]
Use PEMDAS.  So it would then simplify to [100+(586+4)]/5<19 then continueing that it would be [100+590]/5<19 and then 690/5<19 which finally will be 138<19 and that is a valid statement
6 0
3 years ago
A particle moves along a line with acceleration a (t) = -1/(t+2)2 ft/sec2. Find the distance traveled by the particle during the
amm1812

Answer:

s(t)=(ln|7|+ln|2|)\,ft

Step-by-step explanation:

Acceleration is second derivative of distance and are related as:

a(t)=\frac{d^2s}{dt^2}\\\\\frac{d^2s}{dt^2}=\frac{-1}{(t+2)^2}\\

Integrating both sides w.r.to t

v(t)=\frac{ds}{dt}=\frac{1}{t+2} +C\\

Using initial value

v(0)=\frac{1}{2}\\\\\frac{1}{2}=\frac{1}{0+2} +C\\\\C=0\\\\\frac{ds}{dt}=\frac{1}{t+2}

We have to calculate the distance covered in time interval [0,5], so:

\int\limits^5_0 \frac{ds}{dt}=\int\limits^5_0 {\frac{1}{t+2}} \, dt\\\\s(t)=[ln|t+2|]^5_0\\\\s(t)=ln|5+2|+ln|0+2|\\\\s(t)=(ln|7|+ln|2|)\,ft

3 0
3 years ago
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