Find the simple interest paid to the nearest cent for each loan amount, interest rate, and time.

Drag each number to the correct location on the equations. Each number can be used more than once, but not all numbers will be u

Dmitry_Shevchenko [17]

We are given :

The cost of one adult ticket = a.

The cost of one child ticket = c.

Roy buys

6 adult tickets and 2 child tickets.

Total amount = $66.

Therefore, first equation for Roy total amount would be

6a +2c = 66 --------------------(1)

Elisa buys

5 adult tickets and 4 child tickets.

Total amount = $62.

Therefore, second equation for Elisa total amount would be

5a +4c = 62 --------------------(1).

Therefore,

<h3>Roy 6a + 2c = 66</h3><h3>Elisa 5a + 4c = 62.</h3>

4 0

3 years ago

Karen bought a total of seven items at five different stores. She began with $65.00 and had $15.00 remaining. Which of the follo

Zina [86]

In my opinion B would be the best answer

7 0

4 years ago

Principal==12,000 interest is 5.6%..for 5 years.how much interest will I pay.

Keith_Richards [23]

Answer:

Step-by-step explanation:

IF Simple interest; Simple Interest formula would be

S.I = Principal *rate * time

Principal = 12,000

rate = 5.6% or 0.056 as a decimal

time = 5 years

therefore S.I = 12,000 * 0.056 * 5

S.I = 3,360

Therefore, based on simple interest, you will pay $3,360 in interest.

IF Compound interest, the formula would be;

Compound interest = [ Principal *(1+rate )^t ] - Principal

= [ 12,000*(1+0.056)^5 ] -12,000

= [12,000*1.313165883] - 12,000

=15,757.9906 -12,000

= 3,757.9906

Therefore, based on compounding, you will pay $3,757.9906 in interest.

3 0

4 years ago

Which number is a solution to the equation 3x - 5 = 2x + 7? *

igor_vitrenko [27]

Answer:

12

Step-by-step explanation:

4 0

3 years ago

Jamison graphs the function ƒ(x) = x4 − x3 − 19x2 − x − 20 and sees two zeros: −4 and 5. Since this is a polynomial of degree 4

Art [367]

We are given polynomial function f(x)= $x^4-x^3-19x^2-x-20$

We are given two zeros -4 and 5.

Therefore, x=-4 and x=5 is given.

So, the two factors of the polynomial will be (x+4)(x-5).

Let us write some steps to factor the given polynomial.

Let us take above factor (x+4) first.

On factoring (x+4) we get factors

$x^4-x^3-19x^2-x-20=\left(x+4\right)\left(x^3-5x^2+x-5\right)$

$\mathrm{:Let \ us \ factor}\:x^3-5x^2+x-5 \ now.$

Grouping

$=\left(x^3-5x^2\right)+\left(x-5\right)$

Factoring out gcf from each group, we get

$=x^2\left(x-5\right)+\left(x-5\right)$

$=\left(x-5\right)\left(x^2+1\right)$

So, the final factored form of given polynomial will be

$x^4-x^3-19x^2-x-20=\left(x+4\right)\left(x-5\right)\left(x^2+1\right)$

For first to factors (x+4) and (x-5) we have given roots: -4 and 5.

Let us find the root of third factor we got.

$x^2+1=0$

Subtracting both sides by 1.

x^2 = - 1.

On taking square root on both sides, we get a square root(-1)

$x=\sqrt{-1}=$ + i and -i.

Those are imaginary solutions.

Therefore, correct option is B) No, there are two imaginary solutions.

5 0

4 years ago