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qaws [65]
3 years ago
15

How many solutions exist for the system of equations below?

Mathematics
2 answers:
spin [16.1K]3 years ago
6 0

Answer:

none

Step-by-step explanation:

pickupchik [31]3 years ago
3 0
There are no solutions to these two problems.
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klio [65]
2(pi)r

2(pi)0.75
Circumference is 4.71.
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2 years ago
It is predicted that by 2050 there will be 10^10 people living on earth. approximatrly will be 10^12 trees. how many trees for e
Nastasia [14]

Interesting question. Does that mean near the end of humanity?

Solution 1:

10^10 = 1,000,000,000

10^12 = 100,000,000,000

100,000,000,000/1,000,000,000 = 100 trees per person

Solution 2:

10^12 / 10^10  ---  10^10 / 10^10

10^2/1 = 10^2 = 100 trees per person

8 0
3 years ago
Write 4 feet to 15 inches as a ratio. Compare in inches. Write the ratio in lowest terms.
Alja [10]
1 ft = 12 inches....so 4 ft = (4 * 12) = 48 inches

ratio is : 48 to 15...or 48/15 or 48:15......which reduces to 16/5 in lowest terms
8 0
3 years ago
10 PTS!! (5 + 2/8) - (3 7/8)
Artyom0805 [142]
The answer should be 1 3/8
7 0
3 years ago
Read 2 more answers
1 + tanx / 1 + cotx =2
Lera25 [3.4K]

Answer:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

Step-by-step explanation:

Solve for x:

1 + cot(x) + tan(x) = 2

Multiply both sides of 1 + cot(x) + tan(x) = 2 by tan(x):

1 + tan(x) + tan^2(x) = 2 tan(x)

Subtract 2 tan(x) from both sides:

1 - tan(x) + tan^2(x) = 0

Subtract 1 from both sides:

tan^2(x) - tan(x) = -1

Add 1/4 to both sides:

1/4 - tan(x) + tan^2(x) = -3/4

Write the left hand side as a square:

(tan(x) - 1/2)^2 = -3/4

Take the square root of both sides:

tan(x) - 1/2 = (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

tan(x) = 1/2 + (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Take the inverse tangent of both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) = 1/2 - (i sqrt(3))/2

Take the inverse tangent of both sides:

Answer:  x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

4 0
3 years ago
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