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bixtya [17]
2 years ago
8

Solve the system of equations by substitution. Show all work neatly on your work page. Checking your answer is wise. :) Plz need

Mathematics
1 answer:
Flauer [41]2 years ago
7 0

Answer:

x=5

Step-by-step explanation:

3x+2(-3x+11)=7

Use distributive property:

3x-6x+22=7

Combine like terms:

-3x+22=7

-3x=-15

Then divide:

-3x/3=-15/3 = 5

x=5

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3 years ago
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2 years ago
I need some help with these in kinda like a essay form
WINSTONCH [101]

A. The amount (A) at the end of t years of continuous compounding of principal P at rate r will be

... A = Pe^(rt)

For P=1000, r=.02, and t=1, The amount is

... A = $1000e^(.02·1) = $1020.20134

B. The formula for daily compounding is

... A = P(1 + r/365)^(365t)

Using the same values of P, r, t, the amount is

... A = $1000(1 +.02/365)^365 = $1020.20078

Continuous compounding produces a larger result.

The result gets larger the more often compounding occurs. Continuous compounding is the highest possible rate at which compounding can take place, so produces the largest possible result.

C. The balance at the end of the year when interest is compounded n times per year is given by

... A = P(1 + r/n)^n

Each year interest is compounded this way, the amount is multiplied by

... (1 + r/n)^n

When this happens each year for t years, the multiplier has been applied t times. Exponentiation is used to represent the effect of such repeated multiplication, so the balance at the end of t years is

... A = P((1 + r/n)^n)^t = P(1 +r/n)^(nt)

D. (Note the previous answer assumed the existence of this answer.) The same logic as for C above applies for each period that compounding takes place. That is, if compounding occurs n times per year, the interest rate applied for each period is the nominal annual rate r divided by the number of periods n. The multiplier applied to the initial principal amount is

... (1 + r/n)

When than factor is used n times during the year, the multiplier of the initial principal amount is

... (1 + r/n)·(1 + r/n)· ... ·(1 + r/n) . . . where the factor is applied n times.

In more compact notation, this multiplier is

... (1 +r/n)^n

When that multiplier is applied to principal P, the account balance A at the end of the year is ...

... A = P(1 +r/n)^n

7 0
3 years ago
What is the perimeter of a square which has the same area as a circle with circumfrence of 4π
Kruka [31]

Answer:

Perimeter square = 8 sqrt(pi)

Step-by-step explanation:

The perimeter of a square is 4*s

The area of a circle is Area = pi * r^2

The circumference of a circle is C = 2*pi * r

C = 4 pi

4pi = 2*pi * r

r = 2

So the area of the circle is pi * r^2 = pi * 2^2 = 4pi

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s^2 = 4*pi

s = sqrt(4*pi)

s = 2*sqrt(pi)

The perimeter = 4 * 2 * sqrt(pi)

The perimeter = 8 * sqrt(pi)

8 0
2 years ago
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