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3 years ago

What is m∠CAD?

Mathematics

2 answers:

SSSSS [86.1K]3 years ago

8 0

Answer:

100

Step-by-step explanation:

All angles in a triangle equal to 180.

80+20=100

180-100=80

The line is 180

180-80=100

Alex777 [14]3 years ago

4 0

Answer:

100°

Step-by-step explanation:

see attached as reference.

the exterior angle ∠CAD is the sum of its two remote interior angles ∠ACB and ∠CBA

∠CAD = 80 + 20 = 100°

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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 $\frac{dP}{dt}=3.68$ .

7 0

3 years ago

What is the answers to the parts 2a-2d

il63 [147K]

2a) if 1 ft³ weighs 150 lb==>the TOTAL VOLUME =5,000/150 =33.334 ft³

2b) 1 ft³ 1,728 in³. So the TOTAL volume in in³ =33.334 x 1728 = 57,600 in³

c) Volume =(1/3)(πR²).H but R = H then V= (1/3)(πR³).plug V (=57600)

57,600 = 1/3 (πR³) ==> R³ = (3 x 57600) / π ==> R = 38 in

d) Area x thickness = Volume ==> Area x 2 in = 57600 in then:
Are =57600/2 & Area =28,800 in²

5 0

3 years ago

Which of the following is the correct solution to the linear inequality shown below?

Georgia [21]

Correct answer is C.

Line $y=\frac{1}{3}x-2$ intersects y-axis at -2 and passes through the point (3,-1).

Line must be dashed, because the inequality sign is ">".

Point (0,0) must lie in the solution set because it satisfies inequality $y>\frac{1}{3}x-2$ .

$0\ \textgreater \ \frac{1}{3}\times 0-2
\\
\\0\ \textgreater \ -2$

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3 years ago

How do you do recursive formulas

Norma-Jean [14]

Answer:

Determine if the sequence is arithmetic (Do you add, or subtract, the same amount from one term to the next?)

Find the common difference.

Step-by-step explanation:

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2 years ago

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