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dimaraw [331]
3 years ago
7

01 Sam has $3.30 in quarters and dimes, and the total number of coins is 18. How many quarters and how many dimes?

Mathematics
1 answer:
jeka943 years ago
4 0
10 quarters and 8 dimes
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Help asap!!!!!!!!!!!
svlad2 [7]

Answer:

The answer to your question is:

Step-by-step explanation:

12.-

                x² - 6x = 5

                x² - 6x + (3)² = 5 + (3)²

               x² - 6x + 9 = 5 + 9

               (x - 3)² = 14

13.-           x² - 6x = 12

               x² - 6x + (3)² = 12 + (3)²

               x² - 6x + 9 = 12 + 9

              (x - 3)² = 21

              x - 3 = ±√21

             x1 = √21 + 3                     x2 = -√21 + 3

               x1 = 7.58                           x2 = -1.58

14.-           2x² - 24x = - 20            Divide by 2

                x² - 12 x = -10

               x² - 12x + (6)² = - 10 + (6)²

               x² - 12x + 36 = - 10 + 36

              ( x - 6) ² = 26

               x - 6 = ±√26

x1 = √26 + 6                                 x2 = -√26 + 6

x1 = 11.1                                          x2 = 0.9

3 0
3 years ago
A polynomial is factored using algebra tiles. An algebra tile configuration. 0 tiles are in the Factor 1 spot and 0 tiles are in
Harman [31]

Answer:

The factors of the polynomial are:

(x + 1) and (x - 3).

Step-by-step explanation:

4 0
2 years ago
Mara and Jonathan each created a sequence of numbers using a rule. Both sequences start with 0 as the first number. Mara’s rule
Masja [62]

Answer:

Both can be true irrespective of the common difference.

6 0
3 years ago
Prove 2^n > n for all n equal to or greater than 1. I mostly need help with how to solve the problem when it is greater than
noname [10]
If n is an integer, you can use induction. First show the inequality holds for n=1. You have 2^1=2>1, which is true.

Now assume this holds in general for n=k, i.e. that 2^k>k. We want to prove the statement then must hold for n=k+1.

Because 2^k>k, you have

2^{k+1}=2\times2^k>2k

and this must be greater than k+1 for the statement to be true, so we require

2k>k+1

for k>1. Well this is obviously true, because solving the inequality gives 3k>1\implies k>\dfrac13. So you're done.

If you n is any real number, you can use derivatives to show that 2^n increases monotonically and faster than n.
7 0
3 years ago
Can somebody help me and answer this I’ll give u thanks if ur right here’s the picture
sp2606 [1]

Answer:

H

Step-by-step explanation:

The additive inverse property states that x-x=0, or x+(-x). The property that represents this is H.

Hope this helps plz mark brainliest :D

5 0
2 years ago
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