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givi [52]
3 years ago
13

I desperately need the answers to this, it’s about compound interests

Mathematics
1 answer:
Allushta [10]3 years ago
5 0

Answer:

  • quarterly is n = 4
  • semiannually is n = 2
  • monthly is n = 12
  • annually is n = 1

Step-by-step explanation:

n is the number of times the interest is compounded, so just figure out stuff like monthly and annually mean. (ex. since annually means once a year, n = 1)

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Solve for y in terms of x.
krek1111 [17]

<u>Answer:</u>

y=\frac{3}{2} x+6

<u>Step-by-step explanation:</u>

We are given the following equation and we are supposed to solve y in terms of x. It simpler terms, it means that we have to make y the subject of the equation while x being used in it as it is:

\frac{2}{3} y-4=x

Taking the constant 4 to the side where x is to get:

\frac{2}{3} y=x+4

Multiplying the denominator 3 to the other side of the equation to get:

2y=3x+12

Isolating y to make it the subject:

y=\frac{3}{2} x+\frac{12}{2}\\ \\y=\frac{3}{2} x+6

3 0
3 years ago
Amy ate of a pizza and Jack ate of a pizza. Which of the following is true?
pav-90 [236]

Answer:

c

Step-by-step explanation:

. portion of pizza eaten by Amy = portion of pizza eaten by Jack because it doesn't tell us information how much did they ate.

8 0
3 years ago
Dont skip plzzzzzzzzzzzzzzzzzzzzzzzz
natka813 [3]

Answer:

B

Step-by-step explanation:

The slope is rise over run, meaning it is 3/4. The y-intercept is 2, making the equation y = (3/4)x + 2

5 0
3 years ago
Z = -3x +5y<br> Minimum value of z?
NeTakaya

Answer:

minimum value of z is 0

Step-by-step explanation:

if x and y both equal zero, then the value of z is zero, which is your answer.

7 0
3 years ago
Find an explicit solution of the given initial-value problem. (1 + x4) dy + x(1 + 4y2) dx = 0, y(1) = 0
MissTica

Answer:

a solution is 1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4

Step-by-step explanation:

for the equation

(1 + x⁴) dy + x*(1 + 4y²) dx = 0

(1 + x⁴) dy  = - x*(1 + 4y²) dx

[1/(1 + 4y²)] dy = [-x/(1 + x⁴)] dx

∫[1/(1 + 4y²)] dy = ∫[-x/(1 + x⁴)] dx

now to solve each integral

I₁= ∫[1/(1 + 4y²)] dy = 1/2 *tan⁻¹ (2*y) + C₁

I₂=  ∫[-x/(1 + x⁴)] dx

for u= x² → du=x*dx

I₂=  ∫[-x/(1 + x⁴)] dx = -∫[1/(1 + u² )] du = - tan⁻¹ (u) +C₂ =  - tan⁻¹ (x²) +C₂

then

1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) +C

for y(x=1) = 0

1/2 *tan⁻¹ (2*0) = - tan⁻¹ (1²) +C

since tan⁻¹ (1²) for π/4+ π*N and tan⁻¹ (0) for  π*N , we will choose for simplicity N=0 . hen an explicit solution would be

1/2 * 0 = - π/4 + C

C= π/4

therefore

1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4

5 0
4 years ago
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