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marshall27 [118]
2 years ago
9

How to combine like terms

Mathematics
1 answer:
aev [14]2 years ago
8 0

Example: 1x+2y+3z+4a+5x+6y+7z+8+9

We can see that there is more than one number with the variable x, therefore, we say they're ''like terms'' and because of that they can be summed. We do this with all of the other numbers with similar variables. If no numbers with similar variables are left, like 4a, you don't do anything but write them as they are. You can also see that 8 and 9 can also be summed because neither of them has a variable, therefore they're similar.

In this step, you just do the operation with the numbers and keep the same variable.

(1x+5x)+(2y+6y)+(3z+7z)+4a+(8+9)

(6x)+(8y)+(10z)+4a+(17)

since there are not more numbers similar in variables, this operation is done.

6x+8y+10z+4a+17

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0.6 + 15b + 4 = 25.6 all equivelant
Mila [183]

Answer:

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

Step-by-step explanation:

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

3 0
3 years ago
Mai had 5 candy bars. She ate half of one candy bar and decided to distribute the remaining bars between her two sisters and her
brilliants [131]
1 and 2/3 is the answer
7 0
3 years ago
Read 2 more answers
6x^2-16x+5=0<br> A. 8/3<br> B. -3/8<br> C. 5/6<br> D. -5/16
stealth61 [152]
The answer to the question is a

6 0
3 years ago
Jenny and Natalie are selling cheesecakes for a school fundraiser. Customers can buy chocolate cakes and vanilla cakes. Jenny so
IrinaVladis [17]

The cost of 1 chocolate cake is $ 6 and cost of 1 vanilla cake is $ 7

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

Let "c" be the cost of 1 chocolate cake

Let "v" be the cost of 1 vanilla cake

<em><u>Jenny sold 14 chocolate cakes and 5 vanilla cakes for 119 dollars</u></em>

Therefore, we can frame a equation as:

14 x cost of 1 chocolate cake + 5 x cost of 1 vanilla cake = 119

14 \times c + 5 \times v=119

14c + 5v = 119 ------- eqn 1

<em><u>Natalie sold 10 chocolate cakes and 10 vanilla cakes for 130 dollars</u></em>

Therefore, we can frame a equation as:

10 x cost of 1 chocolate cake + 10 x cost of 1 vanilla cake = 130

10 \times c + 10 \times v = 130

10c + 10v = 130 -------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

Multiply eqn 1 by 2

28c + 10v = 238 ------ eqn 3

<em><u>Subtract eqn 2 from eqn 3</u></em>

28c + 10v = 238

10c + 10v = 130

( - ) --------------------------

18c = 108

c = 6

<em><u>Substitute c = 6 in eqn 1</u></em>

14(6) + 5v = 119

84 + 5v = 119

5v = 119 - 84

5v = 35

v = 7

Thus cost of 1 chocolate cake is $ 6 and cost of 1 vanilla cake is $ 7

8 0
2 years ago
An arithmetic progression is a sequence of numbers in which the distance (or difference) between any two successive numbers if t
earnstyle [38]

Answer:

Step-by-step explanation:

Solution:

- We are to write a program for evaluating the sum to Nth of an arithmetic sequence such that the sequence starts from positive integer 1, 3 , 5 , 7 , .. n.

- The sum to nth for the arithmetic series is given by two parameters i.e first integer a = 1 and the distance between successive integers d = 2 in our case.

- For any general distance d we can write our sum to nth as:

          Sum to nth = a + (a+d) + (a+2*d) + (a+3*d) .... (a + (n-1)*d)

- From above sequence we can see that every successive number is increased by distance d and added in previous answer.

- We will use an iteration loop for a variable "sum", which is cycled by a "range ( , , )" function.

- The parameters of the range functions corresponds to:

                   range ( first integer , last integer , step size )  

                   range ( a , n + 1 , d )

- Then we can cast the loop as follows:

 " int sum = 0

   int d = 2

   int a = 1

      for i in range ( a , n + 1 , d )

            sum += i

  "

- We see that iteration parameter i starts from a = 1, with step size d = 2 and the sum is previously stored sum value plus i for the current loop.

3 0
2 years ago
Read 2 more answers
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