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Alisiya [41]
2 years ago
9

A savings account with an initial deposit of $10,000 has an annual interest rate of 3.2%. The interest on the account is compoun

ded monthly. What is the
balance of the account, to the nearest cent, after 6 years?
Mathematics
1 answer:
makkiz [27]2 years ago
7 0

Answer:

account balance will be $12113.9 and the interest is $2113.9.

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2. You are given the hyperbolic relation modeled by xy = -2. Do the following: a) Rewrite the relation such that the dependent v
N76 [4]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

The given equation is :

  • xy =  - 2

1. The relationship such that dependent variable (y) is isolated is :

  • y =   \dfrac{ - 2}{x}

2. The table accompanying this equation :

  • x = -4 and y = 0.5

  • x = -1 and y = 2

  • x = - 0.5 and y = 4

  • x = 0 and y = undefined

  • x = 0.5 and y = -4

  • x = 1 and y = -2

  • x = 4 and y = -0.5

3. graph of the given equation is in attachment ~

4 0
2 years ago
Answer tis please thanks
harkovskaia [24]

Answer:

12

Step-by-step explanation:

8 0
1 year ago
Read 2 more answers
ABC is isosceles.<br>m &lt; a = 3x + 40 and m &lt; c = x + 50<br>what is m &lt; a ??​
il63 [147K]

Answer:

  • 55°

Step-by-step explanation:

<u>It is assumed the angles A and C are congruent:</u>

  • 3x + 40 = x + 50
  • 3x - x = 50 - 40
  • 2x = 10
  • x = 5

m∠A = 3*5 + 40 = 55°

7 0
3 years ago
Read 2 more answers
This is not math but pls help here are some
devlian [24]

Answer:

There is nothing here do you have a question I would be happy to help!!!:)

Step-by-step explanation:

7 0
2 years ago
These roots of the polynomial equation x^4-4x^3-2x^2+12x+9=0 are 3,-1,-1. Explain why the fourth root must be a real number. Fin
Alex787 [66]

Roots with imaginary parts always occur in conjugate pairs. Three of the four roots are known and they are all real, which means the fourth root must also be real.

Because we know 3 and -1 (multiplicity 2) are both roots, the last root r is such that we can write

x^4-4x^3-2x^2+12x+9=(x-3)(x+1)^2(x-r)

There are a few ways we can go about finding r, but the easiest way would be to consider only the constant term in the expansion of the right hand side. We don't have to actually compute the expansion, because we know by properties of multiplication that the constant term will be (-3)(1)(1)(-r)=3r.

Meanwhile, on the left hand side, we see the constant term is supposed to be 9, which means we have

3r=9\implies r=3

so the missing root is 3.

Other things we could have tried that spring to mind:

- three rounds of division, dividing the quartic polynomial by (x-3), then by (x+1) twice, and noting that the remainder upon each division should be 0

- rational root theorem

4 0
3 years ago
Read 2 more answers
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