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suter [353]
3 years ago
15

Find the area of the sector round to the nearest hundredth.

Mathematics
1 answer:
DIA [1.3K]3 years ago
6 0

Answer:

41.89 units

Step-by-step explanation:

Area of a sector: Θ/360 * (pi)(r)^2

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Solve the Equation: √3x-12=0
konstantin123 [22]

√3x - 12 = 0

√3x = 12

(√3x)^2 = (12)^2

3x = 144

x = 48

7 0
3 years ago
Read 2 more answers
What is the equation of the line that passes through (-2, 2) and (1, -4)?
poizon [28]

Answer:

y = - 2x - 2

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁  = (- 2, 2) and (x₂, y₂ ) = (1, - 4)

m = \frac{-4-2}{1+2} = \frac{-6}{3} = - 2, thus

y = - 2x + c ← is the partial equation

To find c substitute either of the 2 points into the partial equation

Using (- 2, 2), then

2 = 4 + c ⇒ c = 2 - 4 = - 2

y = - 2x - 2 ← equation of line

3 0
3 years ago
Will give brainliest!
irakobra [83]

Answer:

Step-by-step explanation:

Remember that our original exponential formula was y = a b x. You will notice that in these new growth and decay functions, the b value (growth factor) has been replaced either by (1 + r) or by (1 - r). The growth "rate" (r) is determined as b = 1 + r.

An exponential function of a^x (a>0) is always ln(a)*a^x, as a^x can be rewritten in e^(ln(a)*x). By deriving, the term (ln(a)) gets multiplied with a^x. The derivative shows, that the rate of change is similiar to the function itself. For 0<a<1, ln(a) becomes negative and so is the rate of change.

Linear models are used when a phenomenon is changing at a constant rate, and exponential models are used when a phenomenon is changing in a way that is quick at first, then more slowly, or slow at first and then more quickly.

6 0
3 years ago
Not sure if any of this is correct, but it’s what I got so far
Irina18 [472]

Problem 1 is correct. You use the pythagorean theorem to find the hypotenuse.

==================================================

Problem 2 has the correct answer, but one part of the steps is a bit strange. I agree with the 132 ft/sec portion; however, I'm not sure why you wrote \frac{1 \text{ sec}}{132 \text{ ft}}=\frac{0.59\overline{09}}{78 \text{ ft}}*127 \text{ ft}

I would write it as \frac{1\text{ sec}}{132 \text{ ft}}*127 \text{ ft} = \frac{127}{132} \text{ sec} \approx 0.96 \text{ sec}

==================================================

For problem 3, we first need to convert the runner's speed from mph to feet per second.

17.5 \text{ mph} = \frac{17.5 \text{ mi}}{1 \text{ hr}}*\frac{1 \text{ hr}}{60 \text{ min}}*\frac{1 \text{ min}}{60 \text{ sec}}*\frac{5280 \text{ ft}}{1 \text{ mi}} \approx 25.667 \text{ ft per sec}

Since the runner needs to travel 90-12 = 78 ft, this means\text{time} = \frac{\text{distance}}{\text{speed}} \approx \frac{78 \text{ ft}}{25.667 \text{ ft per sec}} \approx 3.039 \text{ sec}

So the runner needs about 3.039 seconds. In problem 2, you calculated that it takes about 0.96 seconds for the ball to go from home to second base. The runner will not beat the throw. The ball gets where it needs to go well before the runner arrives there too.

-------------

The question is now: how much of a lead does the runner need in order to beat the throw?

Well the runner needs to get to second base in under 0.96 seconds.

Let's calculate the distance based on that, and based on the speed we calculated earlier above.

\text{distance} = \text{rate}*\text{time} \approx (25.667 \text{ ft per sec})*(0.96 \text{ sec}) \approx 24.64032 \text{ ft}

This is the distance the runner can travel if the runner only has 0.96 seconds. So the lead needed is 90-24.64032 = 65.35968 feet

This is probably not reasonable considering it's well over halfway (because 65.35968/90 = 0.726 = 72.6%). If the runner is leading over halfway, then the runner is probably already in the running motion and not being stationary.

As you can see, the runner is very unlikely to steal second base. Though of course such events do happen in real life. What may explain this is the reaction time of the catcher may add on just enough time for the runner to steal second base. For this problem however, we aren't considering the reaction time. Also, not all catchers can throw the ball at 90 mph which is quite fast. According to quick research, the MLB says the average catcher speed is about 81.8 mph. This slower throwing speed may account for why stealing second base isn't literally impossible, although it's still fairly difficult.

5 0
3 years ago
Plz help will give brainliest need this asap
Ber [7]

Answer:

plz help and this is fro khan academy if it helps!

Hold on, our servers are swamped. Wait for your answer to fully load.

Step-by-step explanation:

plz help and this is fro khan academy if it helps!

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5 0
2 years ago
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