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liberstina [14]
3 years ago
9

Write an algebraic expression to represent each phrase:

Mathematics
2 answers:
Natalka [10]3 years ago
7 0

Answer: A. 7n+4 B. (n+4)7

Step-by-step: A. Put it in simplest form(7n+4) or your teacher might count it wrong.

B. Make sure to use parentheses to tell them to do it first so when using PEMDAS or GERMDAS they don’t do the multiplication first, but the groups.

sladkih [1.3K]3 years ago
5 0

Answer:

Step-by-step explanation:

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In parallelogram ABCD, AB= x, BC= x+y, CD= y-x and AD= 15. solve for x and y
erma4kov [3.2K]

Answer:

I think 29

I don't know if it's right

7 0
3 years ago
−6−1.5(4b+8)=3(−6−2b)
____ [38]

Answer:

-18 = -18  It is true.

Step-by-step explanation:

-6 -1.5(4b+8) = 3(-6 -2b)

 -6   -6b - 12 = -18 -6b

      -6b -18    = -18 -6b

       +6b                 +6b

            -18   =  -18

              It is true.

4 0
3 years ago
If DF= 9x-39, find EF.
Blizzard [7]

4.333333

Hope this helps

8 0
3 years ago
The legs of a right triangle are 4 units and 7 units. What is the length of the hypotenuse? (4 points) Group of answer choices s
morpeh [17]

Answer:

65 square root

Step-by-step explanation:

I’m smart

5 0
3 years ago
How many solutions does the equation have, x1 + x2 + x3 = 10 , where x1 , x2, and x3 are non-negative integers?
MAVERICK [17]

Non-negative integers are positive integers or zero.

1. When x_1=0, then there are such possible cases for x_2 and x_3:

  • x_2=0,\ x_3=10;
  • x_2=1,\ x_3=9;
  • x_2=2,\ x_3=8;
  • x_2=3,\ x_3=7;
  • x_2=4,\ x_3=6;
  • x_2=5,\ x_3=5;
  • x_2=6,\ x_3=4;
  • x_2=7,\ x_3=3;
  • x_2=8,\ x_3=2;
  • x_2=9,\ x_3=1;
  • x_2=10,\ x_3=0.

In total 11 solutions for x_1=0.

2. For x_1=1, there are such possible cases for x_2 and x_3:

  • x_2=0,\ x_3=9;
  • x_2=1,\ x_3=8;
  • x_2=2,\ x_3=7;
  • x_2=3,\ x_3=6;
  • x_2=4,\ x_3=5;
  • x_2=5,\ x_3=4;
  • x_2=6,\ x_3=3;
  • x_2=7,\ x_3=2;
  • x_2=8,\ x_3=1;
  • x_2=9,\ x_3=0.

In total 10 solutions for x_1=1.

3. This process gives you

  • for x_1=2 - 9 solutions;
  • for x_1=3 - 8 solutions;
  • for x_1=4 - 7 solutions;
  • for x_1=5 - 6 solutions;
  • for x_1=6 - 5 solutions;
  • for x_1=7 - 4 solutions;
  • for x_1=8 - 3 solutions;
  • for x_1=9 - 2 solutions;
  • for x_1=10 - 1 solution.

4. Add all numbers of solutions:

11+10+9+8+7+6+5+4+3+2+1=66.

Answer: there are 66 possible solutions (with non-negative integer variables)

8 0
3 years ago
Read 2 more answers
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