NEED HELP NOW!!!!!!!a 5 inch bamboo shoot doubles in height every 3 days. if the equation y=ab^x , where x is the number of doubling periods , represents the height of the bamboo shoot, what are the values of a and b?

Answers

Answer 1

Answer: Answer a = 5 and b = 2.

Procedure

Initially means x = 0, then

5 = a*b^0 = a*1 = a, then a = 5

When x = 1, the height is the double of 5, i.e 10 and

10 = a.b^1 = a.b ⇒ b = 10/a = 10/5 = 2

The a = 5 and b = 2, so the equation is y =5(2)^x

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Solve the word problem below.Jane took out $135 from her checking account. If Jane now has a balance of $650 in her
account, how much did Jane have in the account before she took out any money?
Step 1: If x represents the variable in the word problem above, explain what the variable should
represent.
Step 2: Write an equation to model the situation.
Step 3: Solve the equation and write your answer in sentence form. Show your work.

Answers

Answer:

x = the amount of money Jane had before she took out any

x - 135 = 650

x = 785

Step-by-step explanation:

Let x = the amount of money Jane had before she took out any

Set up the equation

x - 135 = 650

To solve the equation, we must add 135 on both sides to cancel the -135 on the left side of the equal sign.

x - 135 = 650

x = 785

I hope this helps and please mark me as brainliest!

The equation of the line which passesthrough the point(0,2) and has a gradient
of 1/3 is?

Answers

Answer:

Using point (0,2) and gradient 1/3

Equation of the line is

y - 2 = 1/3(x -0)

Multiply through by 3

You'll get

3y - 6 = x

3y = x - 6 or y = 1/3x - 2

Hope this helps.

(A) this relation is a function

(B) this relation is not a function

Answers

Answer:

(b) this relation is not a function because this relation is a one to many ,functions are only many to one and one to one

Find all relative extrema of the function. use the second derivative test where applicable. (if an answer does not exist, enter dne.) f(x) = x3 − 9x2 + 2

Answers

First find the derivative of the function. The derivative is $f'(x)=3 x^(2) -18x$ . Now set it equal to 0 to find the critical numbers. $0=3 x^(2) -18x$ . Factor to solve for the zeros of the derivative. $0=3x(x-6)$ . So 3x = 0, and x = 0, or x - 6 = 0 and x = 6. We will make a table with values for -∞<x<0, 0<x<6, 6<x<∞. Pick a value within those boundaries for each interval and find the sign, positive or negative, that results from subbing that number into the derivative. If we choose -1 in the first interval f'(-1)=21, so the function is increasing from negative infinity to 0. If we choose 1 in the second interval, f'(1)=-15, so the function is decreasing from 0 to 6. If we choose 10 in the last interval, f'(10)=120, so the function is increasing from 6 to infinity. The points of extrema are found by subbing the critical x values into the original function. We know the function is increasing from negative infinity to 0, so f(0)=2, and our max point is (0, 2). We know the function is decreasing from 6 to infinity, so f(6)=-106, and our min point is (6, -106). I do this instead of the second derivative test, but they both work.

∠1 and ∠2 form a linear pair. If m∠1 = 5x + 9 and m∠2 = 3x + 11, find the measures of both angles.