What are the like term in this problem 3a+a+5

Answers

Answer 1

Answer: The like terms are the variables which are being multiplied by different coefficients.

In this case, the like terms are:

3a, a

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Now, 3a + a = 4a

----------

Therefore:

3a + a + 5

= 4a + 5

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adria is putting down floor tiles in her kitchen.each tile has an area of 1 square foot. Twelve square feet of tile costs $7. adria's kitchen is 9 feet wide and 16 feet long. How much will Adria spend on tiles?

Answers

9*16=144
144*$7=1008

$1008

Laura walks 3/5 mile to school each day. Isaiah's walk to school is 3 times al long as Laura's how far does Isaiah walk to school each day

Answers

1 4/5 miles, because 3/5 × 3/1 equals 9/5, which simplifies to be 1 4/5. Hope this helps!

1 and 4/5 because u do the resipical of three which is 3/1 x 3/5=9/5 which simplify a down to 1 and 4/5

Find the LCM of the numbers using prime factorization of 32 and 48

Answers

Answer:

The answer is 96.

Step-by-step explanation:

Ok so all you have to do is list some multiples of each number.

32: 32, 64, 96, 128, 160, 192

48: 48, 96, 144, 192

As you can see, both numbers have the multiple 96, and that's the first one they have in common.

So the lcm is 96.

An international phone call costs $6.53 for 48 minutes. How many cents per minute

Answers

Answer:

14 cents

Step-by-step explanation:

6.53/48=

13.604166667

about 14 cents

Someone please help me

Answers

22 tables because 22x7 is 154
or you could divide 154 by 7

How many tens are in the number 359

Answers

There are 5 tens in 359.

Let me break it down for you 300+50+9 equals 359 is there are 50 because there is just 50 tens in the number 359

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Plz help me which is equivalent

Which system of equations can be used to solve the following problem?Each child ticket for a ride costs $3, while each adult ticket costs $5. If the ride collected a total of $115, and 33 tickets were sold, how many of each type of ticket were sold? Let c be the number of child tickets and a be the number of adult tickets.A.3c+5a=33a + c=115B. 3a+5c=33a + c=115C. 5c+3a=115c + a=33D. 3c+5a=115a +c

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