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Juliette [100K]
2 years ago
13

Find the value of x. Round to the nearest tenth.

Mathematics
1 answer:
tiny-mole [99]2 years ago
4 0

Answer:

x=18.0

Step-by-step explanation:

x^2=a^2+b^2-2ab\;cos\; (30)

x^2=16^2+30^2-2(16)(30)cos(30)

x^2=324.62

x=18.0

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Solve the equation by taking square roots. Problem: 8b^2-7=-76
yulyashka [42]

Answer:

b = +5

Step-by-step explanation:

8b^2=193+7

8b^2=200

b^2= 200/8

b^2=25

b=√25

b=+5

6 0
3 years ago
Karen has already knit 15 centimeters of scarf, and can knit 1 centimeter each night. After
Tems11 [23]

Answer:

I think it is 38 centimeters.

Step-by-step explanation:

7 0
3 years ago
What is the end behavior of the graph of the polynomial function y = 10x9 – 4x?
scoray [572]

Answer:

See below.

Step-by-step explanation:

The value of the highest degree ( x^9)  is 9 - odd.

So this well rise from negative infinity on the left and rise to positive infinity on the right - or, putting it in a different way, fall to the left and rise to the right.

4 0
3 years ago
Read 2 more answers
The rabbit population of Springfield, Ohio was 144,000 in 2016. It is expected to decrease by about 7.2% per year. Use an expone
ehidna [41]

Answer:

The estimated Rabbit population by the year 2036 is 32,309 rabbits

Step-by-step explanation:

In this question, we are expected to use the exponential decay function to estimate population of rabbits in a certain year.

An exponential decay function refers to an equation that estimates the value of a parameter(dependent parameter) at a certain value of the independent parameter given that the independent parameter decreases at a certain constant rate.

Firstly, what we need to do is to write the decay function. To do this, we shall be representing the population by variable P, the rate by r , the number of years by t and the initial population by I

Mathematically, we have the decay function as;

P = I(1-r)^t

From the question, we identify these values as;

P = 144,000 : r = 7.2% =  7.2/100 = 0.072, I = 144,00 and t = 2036-2016 = 20 years

Let's plug these values;

P = 144,000(1-0.072)^20

P = 144,000(0.928)^20

P=  32,309

5 0
3 years ago
On February 1, 1977, John deposited $2250 into a savings account paying 5.76% interest, compounded quarterly. If he hasn't made
Travka [436]
John's effective annual rate is about
  (1 +.0576/4)^4 -1 ≈ 5.8856%
According to the "rule of 72", John's money will have doubled in
  72/5.8856 = 12.23 years

John's balance will be $4500 in 1989.

_____
Since you're only concerned with the year (not the month), you don't actually need to determine the effective annual rate. The given rate of 5.76% will tell you 72/5.76 = 12.5 years. The actual doubling time is closer to 12.12 years, so using the effective rate gives results that are closer, but "good enough" is good enough in this case.
5 0
3 years ago
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