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2 years ago

Mario transferred a balance of $6050 to a new credit card at the beginning of

Mathematics

2 answers:

Kryger [21]2 years ago

7 0

Answer:

the answer is (6050){1+0.031/12}^3{1+0.206/12}^9 for APEX

Step-by-step explanation:

Maru [420]2 years ago

5 0

Answer:

$6050(1+\frac{0.031}{12})^{3}(1+\frac{0.206}{12})^{9}$

Step-by-step explanation:

p = $6050

The card offered an introductory APR of 3.1% for the first 3 months

r = 3.1% or 0.031

t = 3

n = 12

So, compound interest formula for this period is :

$6050(1+\frac{0.031}{12})^{3}$

And a standard APR of 20.6% thereafter,

r = 20.6% or 0.206

t = 9

n = 12

So, compound interest formula for this period is :

$6050(1+\frac{0.206}{12})^{9}$

Now as the p is common, we can re write the expressions as :

$6050(1+\frac{0.031}{12})^{3}(1+\frac{0.206}{12})^{9}$

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Square root of 3x^2y^3 times 2 square root of 72x^3y^4

galina1969 [7]

I hope this helps you

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3 years ago

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

erma4kov [3.2K]

Answer:

$\frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

Step-by-step explanation:

Given

5 tuples implies that:

$n = 5$

$(h,i,j,k,m)$ implies that:

$r = 5$

Required

How many 5-tuples of integers $(h, i, j, k,m)$ are there such that $n\ge h\ge i\ge j\ge k\ge m\ge 1$

From the question, the order of the integers h, i, j, k and m does not matter. This implies that, we make use of combination to solve this problem.

Also considering that repetition is allowed: This implies that, a number can be repeated in more than 1 location

So, there are n + 4 items to make selection from

The selection becomes:

$^{n}C_r => ^{n + 4}C_5$

$^{n + 4}C_5 = \frac{(n+4)!}{(n+4-5)!5!}$

$^{n + 4}C_5 = \frac{(n+4)!}{(n-1)!5!}$

Expand the numerator

$^{n + 4}C_5 = \frac{(n+4)!(n+3)*(n+2)*(n+1)*n*(n-1)!}{(n-1)!5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5*4*3*2*1}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

Solved

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2 years ago

(57x2:3)+32x450-[39-(4+2):2[+8=

tigry1 [53]

Answer:

57×2/3+32×450-[39-4+2]/2+8

57×2/3+32×450-(39-6/2)+8

57×2/3+32×450-36+8

57×2/3+14400-28

57×2/3+14372

38+14372

14410

8 0

2 years ago

-8y^3( 7y^2- 4y - 11) How do you Factor this?

marshall27 [118]

-8y^3( 7y^2- 4y - 11)

-8y^3 (7y-11) (y+1)

8 0

2 years ago

Substitute the values for a, b, and c into b2 – 4ac to determine the discriminant. Which quadratic equations will have two real

hoa [83]

Answer: $0=2x^2-7x-9\\0=4x^2-3x-1\\0=x^2-2x-8$

Step-by-step explanation:

We know that the standard quadratic equation is ax^2+bx+c=0

Let's compare all the given equation to it and , find discriminant.

1. a=2, b= -7, c=-9

$b^2-4ac=(-7)^2-4(2)(-9)=49+72>0$

So it has 2 real number solutions.

2. a=1, b=-4, c=4

$b^2-4ac=(-4)^2-4(1)(4)=16-16=0$

So it has only 1 real number solution.

3. a=4, b=-3, c=-1

$b^2-4ac=(-3)^2-4(4)(-1)=9+16=25>0$

So it has 2 real number solutions.

4. a=1, b=-2, c=-8

$b^2-4ac=(-2)^2-4(1)(-8)=4+32=36>0$

So it has 2 real number solutions.

5. a=3, b=5, c=3

$b^2-4ac=(5)^2-4(3)(3)=25-36=-9$

Thus it does not has real solutions.

5 0

2 years ago

Read 2 more answers