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PilotLPTM [1.2K]

3 years ago

8

Can someone help me with this ( also pls tell me if u can't see the picture because I've posted this question many times and it

won't show up)

1 answer:

Snowcat [4.5K]3 years ago

7 0

That is an obtuse angle.

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A child is flying a kite 145ft above the ground. The string makes an angle of 57 degrees from the child's hands. How much has be

Arturiano [62]

Answer: the length of string that been let out to fly the kite this high is 172.89 ft

Step-by-step explanation:

The length of string attached to the kite, the vertical height of the kite above the ground and the ground distance forms a right angle triangle.

With an angle of 57 degrees, the length of the string that is attached to the kite represents the hypotenuse of the right angle triangle.

The height of the kite above the ground represents the opposite side of the triangle

To determine h, the length of the string that has been let out to fly the kite this high, we would apply the

Sine trigonometric ratio which is expressed as

Sine θ = opposite side/hypotenuse

Sin 57 = 145/h

h = 145/Sin57 = 145/0.8387

h = 172.89

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3 years ago

6 + (9 − 1) 2 ÷ 4 <br> the 2 is an exponent

zmey [24]

6+(9-1)^2/4 = 22
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3 years ago

(10) Jasmine uses 14.24 pounds of fruit for

solniwko [45]

The answer is 0.89 pounds of fruit per serving- please give brainliest answer!

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3 years ago

12 things that has energy

Lapatulllka [165]

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Step-by-step explanation:

8 0

3 years ago

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Using Newton's Law of Cooling. You are a medical examiner called to an abandoned house in which a dead body has been discovered.

Artyom0805 [142]

A) The differential equation comes from the fact that the rate of temperature change is proportional to the difference in temperatures.
$\frac{dT}{dt} = \alpha(T -52)$

B) Find general solution by separating variables and integrating $\int\frac{dT}{T-52} = \alpha \int dt \\ ln (T-52) = \alpha t + C \\ T = C e^{\alpha t} +52$ :

C) Initial condition is t=0, T = 87
$87 = C +52 \\ C = 35$

D) Total time elapsed is 10 minutes, new temperature is 84
$84 = 35 e^{10\alpha} +52$
solve for alpha
$e^{10\alpha} =\frac{32}{35} \\ \alpha = \frac{ln(32/35)}{10} \\ \alpha = -.00896$

E) Temperature function is:
$T = 35 e^{-.00896 t} +52$
solving for t
$t = \frac{ln (\frac{T-52}{35})}{-.00896}$
Plug in T = 98.6
$t = -31.95$

This is approximately 32 minutes before he arrived or about 1:20 AM

3 0

3 years ago