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dolphi86 [110]
3 years ago
11

elena receives $200 for a birthday present, which she deposited in a bank account. Every month she withdraws $25. how much will

she have left in her bank?
Mathematics
1 answer:
Ivan3 years ago
4 0

Answer:

8 months

Step-by-step explanation:

for every 100 four 25s and these two 100s so 8

hope this helps

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Answer:

6.7-(-2.5)=9.2                  6.7+(-2.5)=4.2                          Neither

2.5 + 6.7                          6.7 - 2.5                                  2.5 - 6.7

                                       -2.5 + 6.7                                 -2.5 - 6.7

                                                                                        -6.7 + 2.5

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3 years ago
Jalyn plans to sell Algebra Nation t-shirts as a fundraiser for a new scholarship initiative. The wholesale t-shirt company char
ivolga24 [154]
This situation can be represented by a step function.If Jalyn purchases 60 shirts the total cost will be $600.At these prices it is cheaper to buy 76 shirts than it is to buy 75 shirts.
7 0
3 years ago
Jason purchased a pack of game cards that was on sale for 13% off. The sales tax in his county is 8%. Let y represent the origin
belka [17]

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Step-by-step explanation:

7 0
2 years ago
Which is the simplified form of the expression ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript nega
AlekseyPX

Answer:

The option "StartFraction 1 Over 3 Superscript 8" is correct

That is \frac{1}{3^8} is correct answer

Therefore [(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}

Step-by-step explanation:

Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared

The given expression can be written as

[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2

To find the simplified form of the given expression :

[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2

=(2^{-2})^{-3}(3^4)^{-3}\times (2^{-3})^2(3^2)^2 ( using the property (ab)^m=a^m.b^m )

=(2^6)(3^{-12})\times (2^{-6})(3^4) ( using the property (a^m)^n=a^{mn}

=(2^6)(2^{-6})(3^{-12})(3^4) ( combining the like powers )

=2^{6-6}3^{-12+4} ( using the property a^m.a^n=a^{m+n} )

=2^03^{-8}

=\frac{1}{3^8} ( using the property a^{-m}=\frac{1}{a^m} )

Therefore [(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}

Therefore option "StartFraction 1 Over 3 Superscript 8" is correct

That is \frac{1}{3^8} is correct answer

6 0
3 years ago
Read 2 more answers
PLZZZZZZZzzz help i dont get it
Anna71 [15]
4/5-3/7. First they need the same denominater: 4/5(7)=28/35, 3/7(5)=15/35. 28/35-15/35. Now subtract numerators: 28-15=13. So your answer H. 13/35 :)
6 0
3 years ago
Read 2 more answers
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