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Sloan [31]
3 years ago
15

Let p=x^2-2 which equation is equivalent to (x^2-2)^2+18=9x^2-18 in terms of p

Mathematics
1 answer:
Bad White [126]3 years ago
6 0

Answer:

The equivalent equation is p^2 + 18 = 9p

Step-by-step explanation:

p is given by the following relation:

p = x^2 - 2

And we are given the following equation:

(x^2 - 2)^2 + 18 = 9x^2 - 18

On the right side, we can simplify. So

(x^2 - 2)^2 + 18 = 9(x^2 - 2)

Replacing x^2 - 2 by p, the equation is:

p^2 + 18 = 9p

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8.876 to the nearest whole number
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So your answer is 9
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Rainbow [258]

Answer:

8 one-dollar bills

3 five-dollar bills

2 ten-dollar bills

Step-by-step explanation:

Let x = # of one-dollar bills, y = # of five-dollar bills, and z = # of ten-dollar bills. Total amount in the wallet is $43, so the first equation would be 1x + 5y + 10z = 43. Next, there are 4 times as many one-dollar bills as ten-dollar bills, so x = 4z. There are 13 bills in total, so x + y + z = 13

x + 5y + 10z = 43

x = 4z

x + y + z = 13

x + 5y + 10z = 43

x + 0y - 4z = 0

x + y + z = 13

5y + 14z = 43

-y - 5z = -13

5y + 14z = 43

-5y - 25z = -65

-11z = -22

z = 2

x = 4z

x = 4*2 = 8

x + y + z = 13

8 + y + 2 = 13

10 + y = 13

y = 3

3 0
2 years ago
What is value of x?<br><br><br><br> Enter your answer in the box.
Kryger [21]

Answer:

A number

Step-by-step explanation:

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Find the value of y for the given value of x.
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The fox population in a certain region has a continuous growth rate of 5% per year. It is estimated that the population in the y
kvasek [131]

Answer:

P(t) = A * (1 + r)^t ;

14,922 ;

Year 2013

Step-by-step explanation:

Given the following :

Continuous growth rate(r) = 5% = 0.05

Population in year 2000 = Initial population (A) = 10,100

Time(t) = period (years since year 2000)

A)

Find a function that models the population,P(t) , after (t) years since year 2000 (i.e. t= 0 for the year 2000).

P(t) = A * (1 + r)^t

Trying out our function for t = year 2000, t =0

P(0) = 10,100 * (1 + 0.05)^0

P(0) = 10,100 * 1.05^0 = 10,100

B.)

Use your function from part (a) to estimate the fox population in the year 2008.

Year 2008, t = 8

P(8) = 10,100 * (1 + 0.05)^8

P(8) = 10,100 * 1. 05^8

P(8) = 10,100 * 1.4774554437890625

= 14922.29

= 14,922

c) Use your function to estimate the year when the fox population will reach over 18,400 foxes. Round t to the nearest whole year, then state the year.

P(t) = A * (1 + r)^t

18400 = 10,100 * (1.05)^t

18400/10100 = 1.05^t

1.8217821 = 1.05^t

1.05^t = 1.8217821

In(1.05^t) = ln(1.8217821)

0.0487901 * t = 0.5998151

t = 0.5998151 / 0.0487901

t = 12.293787

Therefore eit will take 13 years

2000 + 13 = 2013

4 0
4 years ago
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