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Alecsey [184]
4 years ago
5

How do you solve 2x^2+4x=3+3x^2?

Mathematics
1 answer:
allochka39001 [22]4 years ago
5 0

Answer:

5x^2+4x=3

Step-by-step explanation:

You add the like terms: 3x^2 and 2x^2, you get 5x^2. You can't do anything else because there are no more terms that are alike.

So the answer would be 5x^+4x=3

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6. What is the value of x?
givi [52]

Answer:

D. 14

Step-by-step explanation:

a^2+b^2=square of c

a^2+x^2=50

6 0
3 years ago
Tickets to a school production cost $5 for a student ticket and $10 for an adult ticket. A total of 67 tickets were purchased at
vazorg [7]

Answer:

Step-by-step explanation:

Keywords:

System of equations, variables, cost, tickets, adults, children.

For this case we must solve a system of equations with two variables represented by the tickets of students and adults of a school production.

We define the variables according to the given table:

a: Number of tickets sold to adults

c: Amount of tickets sold to children.

We then have the following system of equations:

A + c = 67

10a + 5c =440

From the first equation, we clear the value of the variable c:

C = 67 - a

Answer:

The value that could replace c in the table is:

C = 67 - a

Option C is the answer!

Hope it helped u if yes mark me BRAINLIEST!

Tysm! Plz

9 0
3 years ago
Read 2 more answers
Can someone please help mee?
12345 [234]

Answer:

if m<GFD= 67°, then m<GFE is 113°

6 0
3 years ago
Write a recursive rule for the sequence.<br><br> 3, 8, 13, 18, 23, ...
adoni [48]

Answer:

Step-by-step explanation:

c.d. d=8-3=5

a_{n}=a_{n-1}+5

3 0
3 years ago
Solve the following system. y = x 2 - 9x + 10 and x + y + 5 = 0. Enter the solution with the smaller x value first.
pochemuha
X+y+5=0
y=-x-5
If a solution exists y=y so we can say
x^2-9x+10=-x-5  add x+5 to both sides
x^2-8x+15=0 now factor
x^2-3x-5x+15=0
x(x-3)-5(x-3)
(x-5)(x-3) so x=3 and 5, using y=-x-5
y(3)=-8 and y(5)=-10
So the two solutions are:
(3,-8) and (5,-10)

7 0
4 years ago
Read 2 more answers
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