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Elis [28]
3 years ago
10

You roll a number cube two times. How many outcomes are possible? Please explain work.

Mathematics
1 answer:
MArishka [77]3 years ago
6 0

Answer:

there will be about less times than the cube earns!

Step-by-step explanation:

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Hello, I'm new! I need help
melamori03 [73]

The interest in the first month is given as $ 97.1. The principal balance in the second question is $15,030.02

<h3>How to solve for the interest in the first month</h3>

1. We have to solve for the cost of the car

This would be = 19,725*(1.0475)

= 20,661.9375

There is a Down payment = 2,175

balance would be 20661.9375-2,175 = 18,486.94

average rating interest of new car = 6.30%

So the interest accrued in first month = 18,486.94x0.063/12 = $ 97.1

2. cost = 15867

sales tax = 5.25%

10 percent down payment

5.25/100 = 0.0525

cost of car = 15867 + (15867 * 0.0525)

= 16700 dollars

10% of 16700 dollars

= 1670 dollars

principal balance = 16700 - 1670

= $15,030.02

Read more on interest rate here

brainly.com/question/25793394

7 0
2 years ago
At the market, 5 light bulbs cost $9.85. How much do 7 light bulbs cost?
ser-zykov [4K]
First, find what one lightbulb costs.

9.85 / 5 = 1.97

Then multiply the product by 7.

1.97 * 7 = 13.79

Seven light bulbs will cost $13.79
5 0
3 years ago
Hey can you please help me posted picture of question
zhuklara [117]
For this case we have the following polynomial:
 x2 + 6x + 8
 We note that the polynomial can be rewritten as:
 (x + 4) (x + 2)
 Answer:
 
A common factor binomial for this case is given by:
 
(x + 4)
 
option B
7 0
3 years ago
Read 2 more answers
I need help with number 4 and number 3 please
dangina [55]
For Question 3, 
15      ?
__ = ____
120     100

120?=1500
1500/120=12.5

?=12.5%

Question 4

I think it is $462.80


5 0
3 years ago
Solve each of the following equations using a method other than the Quadratic Formula.
liraira [26]

Answer:

A)

y^2-6y = 0

or, y(y-6) = 0

or, y = 0 or y = 6

B)

n^2+5n+7 = 7

or, n^2+5n+7-7 = 7-7 ( Subtracting 7 from both sides)

or, n^2+5n = 0

or, n(n+5) = 0

or, n=0 or n= -5

C)

2t^2-14t+3 = 3

or, 2t^2-14t = 0

or, 2t(t-7) = 0

or, t=0 or t=7

D)

1/3x^2+3x-4 = -4

or, 1/3x^2+3x = 0

or, 1/3x(x+9) = 0

or, x=0 or x= -9

E)

Zero is a common solution to each of the equations. This is because each of the equations had a variable outside the parenthesis with an operation of multiplication.

THANK YOU FOR READING.

8 0
2 years ago
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