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Ostrovityanka [42]
3 years ago
10

Nolan begins a savings account with $100 at 2% interest according to the equation , where Vn is the value of his account after x

years. Anias invests $100 at 2% interest three years after Nolan. The value of Anias’s account, Va is determined using the equation . If Anias would have invested when Nolan first invested, approximately how much money would she have needed to invest to have the same amount of money that she has three years later? $5.77 $94.00 $94.23 $106.12
Mathematics
1 answer:
AleksAgata [21]3 years ago
6 0
$94.23...............
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I will mark u brainliest please help
musickatia [10]

Answer:

30+(2x+7)+63=180(sum of angles of triangle )

30+2x+7+63=180

2x+100=180

2x=180-100

x=80/2

x=40

6 0
2 years ago
Read 2 more answers
If –y + 2x^2 – x = -5y^3 then find dy / dx in terms of x and y.
vlada-n [284]

Answer:

dy/dx = \frac{4x-1}{1-15y^{2} }

Step-by-step explanation:

-y + 2x²- x = -5y³

2x² - x = y - 5y³

4x - 1 = dy/dx - 15y².dy/dx

dy/dx(1 - 15y²) = 4x - 1

dy/dx = \frac{4x-1}{1-15y^{2} }

8 0
2 years ago
A series of quarterly payments of P1,300 each at the first payment is due at 4 years, and the last payment at the end of 12 year
gtnhenbr [62]

Answer:

The Value of Payments is  P18,557.15

Step-by-step explanation:

The quarterly payment is an annuity payment.

Use the following formula to calculate the present value of the payments.

PV of Annuity = Annuity Payment x ( 1 - ( 1 + interest rate )^-Numbers of periods ) / Interest rates

Where

Annuity Payment = Quarterly payment = P1,300

Interest rate = 5.12% x 3/12 = 1.375%

Numbers of periods = 4 years x 12/3 = 16 quarters

PV of Annuity = Value of Payments  = ?

Placing values in the formula

Value of Payments = P1,300 x ( 1 - ( 1 + 1.375% )^-16 ) / 1.375%

Value of Payments =  P18,557.15

4 0
2 years ago
PLS HELP Find the value of x.<br> A. 4<br> B. 2<br> C. 7<br> D. 14
mars1129 [50]

Answer:

ANSWER IS B

Step-by-step explanation:

5x - 3 = 3x + 1

5x - 3x = 1 + 3

2x = 4

x = 4/2

x = 2

4 0
3 years ago
Read 2 more answers
Consider the following functions. f1(x) = x, f2(x) = x2, f3(x) = 2x − 4x2 g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, a
bekas [8.4K]

9514 1404 393

Answer:

  • (c1, c2, c3) = (-2t, 4t, t) . . . . for any value of t
  • NOT linearly independent

Step-by-step explanation:

We want ...

  c1·f1(x) +c2·f2(x) +c3·f3(x) = g(x) ≡ 0

Substituting for the fn function values, we have ...

  c1·x +c2·x² +c3·(2x -4x²) ≡ 0

This resolves to two equations:

  x(c1 +2c3) = 0

  x²(c2 -4c3) = 0

These have an infinite set of solutions:

  c1 = -2c3

  c2 = 4c3

Then for any parameter t, including the "trivial" t=0, ...

  (c1, c2, c3) = (-2t, 4t, t)

__

f1, f2, f3 are NOT linearly independent. (If they were, there would be only one solution making g(x) ≡ 0.)

7 0
2 years ago
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