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3 years ago

How tall is the tree? A. 8ft B. 15 Ft C. 18 FT D. 72 ft

Mathematics

2 answers:

Kobotan [32]3 years ago

8 0

We are supposed to assume ratio of similar triangles (since he is looking at mirror, angle of incendence=angle of reflection so all 3 angles are same and the triangles are similar)
use ratio
ratio of height of small triangle to base of small triangle=ratio of height of large triangle to base of large triangle
so6:2=h:246/2=h/243=h/24multiply both sides by 2472=hthe tree is 72ft tall
answer is D

Nostrana [21]3 years ago

5 0

The answer is d multiple the andeers

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yuradex [85]

Answer:

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Step-by-step explanation:

Two coordinate points are (-3,-3) and (3,1). If you use Y2-Y1/X2-X1, you will find the slope. Then use y=mx+b to find the equation.

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2 years ago

The sides of a rectangle are x cm and 4 cm. Its<br>perimeter is 46 cm. Determine the value of x.

Vedmedyk [2.9K]

Answer:

x = 19 cm

Step-by-step explanation:

perimeter = 46 = 2(x + 4)

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3 years ago

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tankabanditka [31]

Answer:

30.96 m^2

Step-by-step explanation:

Side length of the square is also the diameter of the semicircle so:

r = 0.5 * 12 m = 6 m

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3 years ago

A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per

Tems11 [23]

Answer:

The pressure is changing at $\frac{dP}{dt}=3.68$

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ and we want to find at what rate is the pressure changing.

The equation that model this situation is

$PV^{1.4}=k$

Differentiate both sides with respect to time t.

$\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\$

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

$\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$

Apply this rule to our expression we get

$V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0$

Solve for $\frac{dP}{dt}$

$V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}$

when P = 23 kg/cm2, V = 35 cm3, and $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ this becomes

$\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68$

The pressure is changing at $\frac{dP}{dt}=3.68$ .

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3 years ago

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Rashid [163]

You could do it 3 ways.
Hope this helps.

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3 years ago

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