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ch4aika [34]
3 years ago
15

Can someone help me with these two problems please and thank you I need them asap

Mathematics
1 answer:
BlackZzzverrR [31]3 years ago
8 0
Can please send a bigger pic if you really want someone to solve that problem
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Why would a bank charge a fee to non-customers to use their ATM?
worty [1.4K]

Answer:

Charging a noncustomer helps pay for the cost of keeping the ATM running and supplied with cash. Customers of the bank already cover the ATM costs with regular fees or by allowing the bank to loan out their money.

Step-by-step explanation:

8 0
3 years ago
What is the result of adding these two equations?<br> 4x – 4y = -2<br> –9x – 4y = -3
White raven [17]
The answer to your question 4x-4y= -2 plus -9x-4y= -3 is -5 because in order to find the result of the two equations would be to add the solutions to both of them together.
7 0
3 years ago
Solve for y<br> 2(3 + 3y) + y = 11
garik1379 [7]
<h2><u>EQUATION</u></h2><h3>Exercise</h3>

2(3 + 3y) + y = 11

First, apply the distributive property:

2(3 + 3y) + y = 11

6 + 6y + y = 11

6 + 7y = 11

Substract 6 from both sides:

6 - 6 + 7y = 11 - 6

7y = 5

Divide both sides by 7:

\dfrac{7y}{7} = \dfrac{5}{7}

\boxed{y = \dfrac{5}{7}}

<h3><u>Answer</u>. The value of y = 5/7.</h3>

8 0
3 years ago
In how many ways can you answer the question on an exam that consists of 9 questions with 4 answers
lisabon 2012 [21]
I think the answer is 36 I hope I helped
6 0
3 years ago
Express the product in simplest form.
Dmitry [639]

The product in simplest form is (x - 4)

<em><u>Solution:</u></em>

<em><u>Given expression is:</u></em>

\frac{8}{2x+8} \times \frac{x^2-16}{4}

We have to find the product in simplest form

In the given expression,

2x + 8 = 2(x+ 4)

We know that,

a^2-b^2=(a+b)(a-b)

Therefore,

x^2-16 = x^2-4^2 =(x+4)(x-4)

Substitute these in given expression

\frac{8}{2(x+4)} \times \frac{(x+4)(x-4)}{4}

Cancel the common factors,

\frac{8}{2(x+4)} \times \frac{(x+4)(x-4)}{4} = x - 4

Thus the product in simplest form is (x - 4)

8 0
3 years ago
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