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2 years ago

Determine the effective monthly growth rate for a population that grows at a rate of 23% each year

Mathematics

2 answers:

____ [38]2 years ago

7 0

23/12=1.9%..................................

melisa1 [442]2 years ago

6 0

23/12=1.9% the math is easy

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Plzzz help I don’t want to have a bad grade

blondinia [14]

Answer:

B 46.2

Step-by-step explanation:

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2 years ago

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Rachel tutors English. for each hour that she tutors, she earns 50 dollars. Let E be her earnings (in dollars) after tutoring fo

n200080 [17]

Answer:

a. E = 50 H b. 950 dollars

Step-by-step explanation:

a. Since Rachel tutors English for 50 dollars for each hours, her rate is 50 dollar per hour. If she tutors for time, H hours, Her earning E = rate × time = 50 × H

E = 50H.

b. If Rachel's earnings after tutoring for 19 hours is gotten by substituting H = 19 into the equation for the earnings, E.

So, E = 50H

E = 50 × 19

E = 950 dollars.

So, Rachel's earnings after 19 hours is 950 dollars.

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2 years ago

This is due at 11:59 and urgent please answer this, tyyy :))

serious [3.7K]

Answer:

(0,-5). y=mx+b or -5=0 × 0+b, or solving for b: b=-5-(0)(0). b=-5.

(2,-5). y=mx+b or -5=0 × 2+b, or solving for b: b=-5-(0)(2). b=-5.

y=-5

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2 years ago

Polynomial Warm Up

Kazeer [188]

Answer:

Step-by-step explanation:

False you must multiply the coefficient (numbers) and add the exponents.

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2 years ago

(1+cos2x)/(1-cos2x) = cot^2x

sesenic [268]

We will turn the left side into the right side.

$\dfrac{1 + \cos 2x}{1 - \cos2x} = \cot^2 x$

Use the identity:

$\cos 2x = \cos^2 x - \sin^2 x$

$\dfrac{1 + \cos^2 x - \sin^2 x}{1 - ( \cos^2 x - \sin^2 x)} = \cot^2 x$

$\dfrac{1 - \sin^2 x + \cos^2 x }{1 - \cos^2 x + \sin^2 x} = \cot^2 x$

Now use the identity

$\sin^2 x + \cos^2 x = 1$ solved for sin^2 x and for cos^2 x.

$\dfrac{\cos^2 x + \cos^2 x }{\sin^2 x + \sin^2 x} = \cot^2 x$

$\dfrac{2\cos^2 x}{2\sin^2 x} = \cot^2 x$

$\dfrac{\cos^2 x}{\sin^2 x} = \cot^2 x$

$\cot^2 x = \cot^2 x$

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2 years ago