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Mandarinka [93]
3 years ago
14

Kate has 21 coins (nickels and dimes) in her purse. How many nickels and dimes does she have if she has $1.50?

Mathematics
2 answers:
Gre4nikov [31]3 years ago
8 0
She would have 9 dimes and 12 nickels
Lunna [17]3 years ago
4 0
She should have 20 nickels and 5 dimes
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Solve for x in the following system<br><br> 6x + 6y = -2<br> y = -x
klasskru [66]
6x + 6(-x) = - 2
6x - 6x = - 2

the answer was not relevant than 0

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8 0
3 years ago
Keith has $500 in his savings account at the beginning of the summer. He
bixtya [17]

Answer:

12

weeks

Explanation:

If Keith starts with

$

500

and wants to end with (at least)

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200

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−

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=

$

300

If he withdraws

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25

week

the

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300

$

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Step-by-step explanation:

I apologize if this isn't correct, I tried

7 0
3 years ago
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1. Evaluate: 5 ÷1 + 3 + 7​
Elena-2011 [213]

Answer:

15

Step-by-step explanation:

5÷1+3+7

PEMDAS

Parenthesis

Exponents

Multiplication

Division

Addition

Subtraction

5÷1= 5, then 5+3=8, then 8+7= 15

3 0
3 years ago
Another one help please
guapka [62]

there must be subtraction of multiplication of each two diagonal,

|G|=[12×2-(-6)×0]

=(24+0)

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8 0
4 years ago
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Factor the expression using the two different techniques listed for Parts 1(a) and 1(b).
Natalija [7]

Answer:

9a^4 b^10(2 +3a^6 b^5)  (2 -3a^6 b^5)

see work below

Step-by-step explanation:

36a^4b^10 - 81a^16b^20

A)  find the GCF

36a^4b^10 = 4*9 a^4b^10  = 2*2*3*3 a*a*a*a*b*b*b*b*b*b*b*b*b*b

81a^16b^20= 9*9a^16b^20= 3*3*3*3* a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a                *b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b

The terms that appear in both terms is the GCF.  The terms that remain are inside the parentheses.

The GCF is 3*3 a*a*a*a*b*b*b*b*b*b*b*b*b*b

36a^4b^10 - 81a^16b^20 = 3*3 a*a*a*a*b*b*b*b*b*b*b*b*b*b *

( 2*2 - 3*3a*a*a*a*a*a*a*a*a*a*a*a b*b*b*b*b*b*b*b*b*b)

Combining like terms

36a^4b^10 - 81a^16b^20 = 9a^4b^10(4-9a^12b^10)

The expression inside the parenthesis can be factored using the difference of squares

let x^2 =4   x =2  

y^2 = 9a^12 b^10   y = 3a^6b^5

(x^2 -y^2) = (x+y)(x-y)

9a^4b^10(4-9a^12b^10) = 9a^4b^10 ( 2+3a^6b^5) ( 2-3a^6b^5)

b) difference of squares  a^2 – b^2 = (a + b)(a – b)

let a^2 = 36a^4b^10

so a = 6a^2b^5

b^2 = 81a^16b^20

b = 9a^8 b^10

a^2 – b^2 = (a + b)(a – b)

36a^4b^10 - 81a^16b^20 = (6a^2b^5 +9a^8 b^10) (6a^2b^5 -9a^8 b^10)

We can factor a 3 a^2 b^5 out of the first term

3 a^2 b^5 (2 +3a^6 b^5) (6a^2b^5 -9a^8 b^10)

3 a^2 b^5 (2 +3a^6 b^5) 3 a^2 b^5 (2 -3a^6 b^5)

Multiply the terms outside the parentheses together

9a^4 b^10(2 +3a^6 b^5)  (2 -3a^6 b^5)

4 0
3 years ago
Read 2 more answers
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