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4 years ago

14

Peter deposited $27 into his savings account bringing the total to $234.wrotw and solve an inequality to represent the amount of

money in peters account before the $27 deposit

2 answers:

Akimi4 [234]4 years ago

6 0

Before deposit, his account value was $234-$27 =$207

So, inequality would be: 207 > 0

207 +x > 0

where x=deposition money, $27.

Kamila [148]4 years ago

5 0

Your total is $207 that was in the account before he added the $27.

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ki77a [65]

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3 years ago

Read 2 more answers

I have 54 coins in my pocket; some of them are quarters and the rest are nickels. In total, the coins are worth $6.10. How many

ss7ja [257]

Answer:

17 quarters, 37 nickels

Step-by-step explanation:

We just have to set up a system of equations

quarters are x

and nickels are y

x+y=54

0.25x+0.05y=6.10

now we can get ready to substitute

x=54-y

substitute it in

0.25(54-y)+0.05y=6.10

simplify/evaluate

13.5-0.25y+0.05y=6.10

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plug it back in

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3 0

3 years ago

6/7 divided by 2 <br><br> IN FRACTION<br><br> EXPLAIN HOW U GOT

gladu [14]

Answer:

3/7

Step-by-step explanation:

First, when dividing 6/7 by 2 you need to change the number 2 into a fraction. The number 2 as a fraction is 2/1. However, this is division so you need to reverse the fraction so it is 1/2 instead of 2/1. So, 6/7 x 1/2 = 6/14. You can simplify to get your the answer.

Hope this helped!

4 0

3 years ago

What's the answer ? I been having trouble finding the answer

Fiesta28 [93]

What's the answer to what?

5 0

3 years ago

F:x→(x+1/x+3), x≠3 and g:x→（6/x-2), x≠2, find f²(4) and g²(1/2)

Sveta_85 [38]

Answer:

$f^2(4) = (\frac{25}{49})\\g^2(\frac{1}{2} ) = 16$

Step-by-step explanation:

Here, given:

$f(x) = (\frac{x+1}{x+3} ), g(x) = (\frac{6}{x-2} )$

Now, here to find the values of $f^2(4), g^2(\frac{1}{2})$

As we know: $f^n(x) = (f(x))^n$

Now, substituting x = 4 in f(x):

$f(4) = \frac{4 +1}{4+3} = \frac{5}{7} \\\implies f^4 = (f(4))^2 = (\frac{5}{7} )^2 = \frac{25}{49} \\\implirs f^{(4)} = ( \frac{25}{49})$

Now, substituting x = 1/2 in g(x):

$g(\frac{1}{2} ) = \frac{6}{\frac{1}{2}-2} = \frac{6}{-\frac{3}{2}} \\\\ g(\frac{1}{2} ) = -6\times (\frac{2}{3} ) = -4\\\implies g^2{(\frac{1}{2}) = ( g(\frac{1}{2} ))^2 = (-4 )^2 =16 \\\\\implies g^2{(\frac{1}{2}) = 16$

3 0

3 years ago