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Ber [7]
3 years ago
13

4) John has a balance of $1,500 in his First State Bank checking account. He

Mathematics
1 answer:
Anni [7]3 years ago
4 0

Answer:

2036.81

Step-by-step explanation:

1500+477.29+237.52+550=2764.81

2764.81-728=2036.81

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CAN SOMEBODY JUST ANSWER THE CORRECT ONE COME ON
Liono4ka [1.6K]

Answer:

(n-2)180

Step-by-step explanation:

6 0
2 years ago
Karen paid $62 for pair of shoes. Three days later she noticed the shoes were on sale for $50. What was the percent of change
natta225 [31]

Answer:

24%

Step-by-step explanation: c:

7 0
3 years ago
Prove that:
photoshop1234 [79]
(a) 5^31-5^29 = 5^29(5^2-1)

 = 5^29(24)
 = 5^29(6)(4)
 =5^26(6)*[(5)(5)(4)]
 =5^26(6)*100
  (b) 25^9+5^17 = 5^18+5^17 
=5^17(5+1)
 =5^17(6)
 =5^16[(5)(6)]
 =5^16*30 
 (c) 27^10–9^14 = 3^30-3^28
 =3^28(3^2 -1)
 =3^28(8)
 =3^27[(3)(8)]
 =3^27*24

6 0
3 years ago
Can u express 9 and 9 as a mixed number ? Why or why not ? 9and 9 means fraction
Lostsunrise [7]
Yes:

1 & 0/9
   or
1 & 1/1
6 0
4 years ago
1. f(x) = 3x - 2<br> Domain: (-2, 0,4)
kifflom [539]

Step-by-step explanation:

Since the calculations are very easy, I'll just focus on what you actually have to ask yourself to solve the exercise.

Domain: the question you have to ask yourself is "which numbers my function will accept as an input?" or, equivalently, "which numbers my function will not accept as an input?"

From the second question, we know that there are some functions with domain issues: for example, if there is a denominator, you must be sure that it isn't zero, since you can't divide by zero. So, that function wouldn't accept as input the values which annihilate the denominator.

In general, you have domain issues with:

Denominator (cannot be zero);

Even roots (they can't be computed for negative numbers);

Logarithms (they can't be computed for negative numbers, or zero).

Is this case, you have none of the three above, and so you have no domain issues. Alternatively, you could just see that your function picks a number

x

, multiplies it by

3

, and then adds

2

, and of course you can multiply any number by

3

, and you can add

2

to any number.

Range: now you should ask: which values can I obtain from my functions? I say that you can obtain every possible value. Let's say that you want to obtain a particular number

y

. So, you need to find a number

x

such that

3

x

+

2

=

y

, and the equation easily solves for

x

, with

x

=

y

−

2

3

.

So, if you choose any number

y

, I can tell you that it is the image of a particular

x

, namely

y

−

2

3

, and again, this algorithm is ok for any

y

, you simply need to subtract

2

and then divide the whole thing by

3

, which again are operations you are always allowed to do.

3 0
3 years ago
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