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skad [1K]
3 years ago
15

What is 15,409 in word form?

Mathematics
2 answers:
Rudik [331]3 years ago
8 0
Fifteen thousand four hundred nine
polet [3.4K]3 years ago
3 0

fifteen thousand four hundred nine

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Factor the following Polynomial a) 6x - 18​
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U could easily do this on Photo-math- but here: 6(x−3)
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In 2013, the moose population in a park was measured to be 5,100. By 2018, the population was measured again to be 5,200. If the
natka813 [3]

Answer:

P(t) = 5100e^{0.0039t}

Step-by-step explanation:

The exponential model for the population in t years after 2013 is given by:

P(t) = P(0)e^{rt}

In which P(0) is the population in 2013 and r is the growth rate.

In 2013, the moose population in a park was measured to be 5,100

This means that P(0) = 5100

So

P(t) = 5100e^{rt}

By 2018, the population was measured again to be 5,200.

2018 is 2018-2013 = 5 years after 2013.

So this means that P(5) = 5200.

We use this to find r.

P(t) = 5100e^{rt}

5200 = 5100e^{5r}

e^{5r} = \frac{52}{51}

\ln{e^{5r}} = \ln{\frac{52}{51}}

5r = \ln{\frac{52}{51}}

r = \frac{\ln{\frac{52}{51}}}{5}

r = 0.0039

So the equation for the moose population is:

P(t) = 5100e^{0.0039t}

5 0
3 years ago
Rosalie’s employer pays 75% of her health insurance premium and deducts the remainder from her paycheck. Rosalie is paid biweekl
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D is the right answer
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A ball is thrown straight out at 80 feet per second from an upstairs window that's 15 feet off the ground. Find the ball's horiz
Alex73 [517]

Answer:

Step-by-step explanation:

In order to find the horizontal distance the ball travels, we need to know first how long it took to hit the ground. We will find that time in the y-dimension, and then use that time in the x-dimension, which is the dimension in question when we talk about horizontal distance. Here's what we know in the y-dimension:

a = -32 ft/s/s

v₀ = 0 (since the ball is being thrown straight out the window, the angle is 0 degrees, which translates to no upwards velocity at all)

Δx = -15 feet (negative because the ball lands 15 feet below the point from which it drops)

t = ?? sec.

The equation we will use is the one for displacement:

Δx = v_0t+\frac{1}{2}at^2 and filling in:

-15=(0)t+\frac{1}{2}(-32)t^2 which simplifies down to

-15=-16t^2 so

t=\sqrt{\frac{-15}{-16} } so

t = .968 sec (That is not the correct number of sig fig's but if I use the correct number, the answer doesn't come out to be one of the choices given. So I deviate from the rules a bit here out of necessity.)

Now we use that time in the x-dimension. Here's what we know in that dimension specifically:

a = 0 (acceleration in this dimension is always 0)

v₀ = 80 ft/sec

t = .968 sec

Δx = ?? feet

We use the equation for displacement again, and filling in what we know in this dimension:

Δx = (80)(.968) +(0)(.968)^2 and of course the portion of that after the plus sign goes to 0, leaving us with simply:

Δx = (80)(.968)

Δx = 77.46 feet

5 0
3 years ago
682 rounded to the nearest hundred
defon
700, because 8 is 5 and up so it'll be rounded up.
4 0
4 years ago
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