All categories

2 years ago

PLEASE ANSWER ASAP ASSIGNMENT IS DUE TODAY I WILL MARK BRAINLIST!!!!!!!!!!!!

Mathematics

1 answer:

Gnom [1K]2 years ago

4 0

9514 1404 393

Answer:

V(t) = 25(1.07^t)
V(t) = 40(1.05^t)

Step-by-step explanation:

Put the given numbers into the given formula. It can be simplified a little bit.

Formula:

V(t) = P(1 + r)^t

Coin A

P = 25, r = .07

V(t) = 25(1.07^t)

Coin B

P = 40, r = 0.05

V(t) = 40(1.05^t)

You might be interested in

PLZ HELP !

max2010maxim [7]

Answer:

13 cm

Step-by-step explanation:

The diagonal of a rectangle forms a right triangle; where the diagonal is a hypotenuse, and two sides of the rectangle are legs.

Using pythagorean's theorem ( $a^2 + b^2 = c^2$ ), we can say:

$5^2 + 12^2 = c^2\\25 + 144 = c^2\\169 = c^2\\13 = c$

The hypotenuse is equal to 13cm.

3 0

3 years ago

For the line perpendicular to y = 12, the slope is undefined because the denominator of the slope is 0. For the line parallel to

ASHA 777 [7]

I think the answer is b

7 0

2 years ago

a man deposits 45 into an account the next day he makes 3 deposits of 20 $ each a week later he withdraws 105$

zheka24 [161]

Answer:

The amount available in account at last is $0

Step-by-step explanation:

Given as :

The withdrawal amount from the account at the end = $105

The first deposit amount = $45

The next three deposit amount = $20 each

I.e The second deposit amount = 3 × $20

Or, The second deposit amount = $60

Let The amount available in account at last = $A

Now, According to question

The amount available in account at last = (first deposit amount + second deposit amount) - withdrawal amount from the account at the end

i.e A = ($45 + $60) - $105

Or, A = $105 - $105

∴ A = $0

So, The amount available in account at last = A = $0

Hence, The amount available in account at last is $0 . Answer

8 0

3 years ago

What is the slope of the line described by y=3x-6

aleksandrvk [35]

Answer:

The Slope: m=3

6 0

2 years ago

Read 2 more answers

3 determine the highest real root of f (x) = x3− 6x2 + 11x − 6.1: (a) graphically. (b) using the newton-raphson method (three it

Juliette [100K]

(a) See the first attachment for a graph. This graphing calculator displays roots to 3 decimal places. (The third attachment shows a different graphing calculator and 10 significant digits.)

(b) In the table of the first attachment, the column headed by g(x) gives iterations of Newton's Method. (For Newton's method, it is convenient to let the calculator's derivative function compute the derivative f'(x) of the function f(x). We have defined g(x) = x - f(x)/f'(x).) The result of the 3rd iteration is ...

... x ≈ 3.0473167

(c) The function h(x₁, x₂) computes iterations using the secant method. The results for three iterations of that method are shown below the table in the attachment. The result of the 3rd iteration is ...

... x ≈ 3.2291234

(d) The function h(x, x+0.01) computes the modified secant method as required by the problem statement. The result of the 3rd iteration is ...

... x ≈ 3.0477377

(e) Using Mathematica, the roots are found to be as shown in the second attachment. The highest root is about ...

... x ≈ 3.0466805180

_____

Comment on these methods

Newton's method can have convergence problems if the starting point is not sufficiently close to the root. A graphing calculator that gives a 3-digit approximation (or better) can help avoid this issue. For the calculator used here, the output of "g(x)" is computed even as the input is typed, so one can simply copy the function output to the input to get a 12-significant digit approximation of the root as fast as you can type it.

The "modified" secant method is a variation of the secant method that does not require two values of the function to start with. Instead, it uses a value of x that is "close" to the one given. For our purpose here, we can use the same h(x1, x2) for both methods, with a different x2 for the modified method.

We have defined h(x1, x2) = x1 - f(x1)(f(x1)-f(x2))/(x1 -x2).

6 0

2 years ago