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3 years ago

Describe in words how the temperature changes with time, where h is hours since midnight. T=50-2h

Mathematics

1 answer:

Westkost [7]3 years ago

6 0

Answer:

Step-by-step explanation:

So we know h is hours, what is T? Obviously temperature. Let's try plugging in a few values. at h=0, or hours since midnight, which is a fancy way of saying midnight, T = 50, so that meeans it starts at 50 degrees. What about at 1 hour after midnight? or h=1 (or 1 AM) well, T = 50-2(1) = 48. so 48 degrees. You can keep doing this, and what happens?

Well T keeps going down by 2 every time h goes up by 1. Well if T is temperature and h is hours since 12 AM we just replace the words. Temperature keeps going down by 2 (degrees) every time hours since midnight goes up by 1. You can probably write that a bit more well worded, but I wanted to show how you can get the answer from seeing what happens to the variables. Though let me know if you don't get it.

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

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4 years ago

Please help asap i’m stupid

nevsk [136]

Answer:

Step-by-step explanation:

Say that arc JL going through M is arc E and JL going the other way is arc D

For the angle formed by two tangents, K=(1/2)(E-D)

64=E-D

Furthermore, angle K and central angle JCL (facing toward K) are supplementary, so 180-K=JCL=180-32=148

Thus, as the angles around angle C add up to 360, angle JCL (facing toward M) is 360-148=32+180=212

E is then 212

64=212-D

212-64=D=148

Thus, as JML is an inscribed angle, M=1/2(D)=1/2(148)=74

3 0

4 years ago

Describe in words how the temperature changes with time, where h is hours since midnight. T=50-2h​

Describe in words how the temperature changes with time, where h is hours since midnight. T=50-2h