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inessss [21]
4 years ago
7

The amount of grain in a silo is changing at a rate of –35.2 kilograms per hour.

Mathematics
1 answer:
Ainat [17]4 years ago
4 0
Change=change per hour times number of hours
change per hour=-35.2
hours=1.25
change=-35.2 times 1.25
change=-44


answer is
hmm
the change is 44, but it is in a negative direction

either 44 or -44, not sure
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A peregrine falcon can fly 322 kilometers per hour. How many meters per hour can the falcon fly
lianna [129]
322kmh x1000= 322,000 meters per hour
That is the relationship between km and m
Because we are turning bigger units into smaller we the number will increase.

example   2 wks = 14 days  two became 14 - it increased
7 0
3 years ago
1.A house worth $250,000 has an coinsurance clause of 90 percent. The owners insure the property for $191,250. They then have a
Arisa [49]

1. Multiply the value of the house by the percent:

250,000 x 0.9 = 225,000

Divide the amount the owner insured by the clause amount:

191,250/225,000 = 0.85

Multiply that by the amount of damage:

0.85 x 80,000 = 68,000

They will receive $68,000

2. Follow the same steps as above:

180,000 x 0.75 = 135,000

101,250 / 135,000 = 0.75

0.75 x 50,000 = 37,500

They will receive $37,500

8 0
3 years ago
Read 2 more answers
(−3 + 8i)(−3 − 8i) =
Mice21 [21]

Answer: 73

Step-by-step explanation:  

Multiply using the FOIL method, then combine the real and imaginary parts of the expression.

73

8 0
3 years ago
23. What is the angle of rotation of the following figure?<br><br> 45<br> 180<br> 60<br> 90
MA_775_DIABLO [31]

I believe that it is 45 degrees. i could be wrong, so make sure to check!

5 0
3 years ago
Read 2 more answers
How do you do these two questions?
nignag [31]

Step-by-step explanation:

(a) ∫₋ₒₒ°° f(x) dx

We can split this into three integrals:

= ∫₋ₒₒ⁻¹ f(x) dx + ∫₋₁¹ f(x) dx + ∫₁°° f(x) dx

Since the function is even (symmetrical about the y-axis), we can further simplify this as:

= ∫₋₁¹ f(x) dx + 2 ∫₁°° f(x) dx

The first integral is finite, so it converges.

For the second integral, we can use comparison test.

g(x) = e^(-½ x) is greater than f(x) = e^(-½ x²) for all x greater than 1.

We can show that g(x) converges:

∫₁°° e^(-½ x) dx = -2 e^(-½ x) |₁°° = -2 e^(-∞) − -2 e^(-½) = 0 + 2e^(-½).

Therefore, the smaller function f(x) also converges.

(b) The width of the intervals is:

Δx = (3 − -3) / 6 = 1

Evaluating the function at the beginning and end of each interval:

f(-3) = e^(-9/2)

f(-2) = e^(-2)

f(-1) = e^(-1/2)

f(0) = 1

f(1) = e^(-1/2)

f(2) = e^(-2)

f(3) = e^(-9/2)

Apply Simpson's rule:

S = Δx/3 [f(-3) + 4f(-2) + 2f(-1) + 4f(0) + 2f(1) + 4f(2) + f(3)]

S ≈ 2.5103

5 0
3 years ago
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