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Alex17521 [72]
3 years ago
5

How do I solve this question

Mathematics
1 answer:
sammy [17]3 years ago
4 0
The picture is kind of blurry could you please take a clearer one?

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What is the solution to the following question
Anastaziya [24]

Answer:

3

Step-by-step explanation:

2(x + 1) = 3(5 - x) + 2

Distribute,

2(x + 1) = 3(5 - x) + 2

2x + 2 = 3(5 - x) + 2

Rearrange terms,

2x + 2 = 3(5 - x) + 2

2x + 2 = 3(-x + 5) + 2

Distribute,

2x + 2 = 3(-x + 5) + 2

2x + 2 = -3x + 15 + 2

Add the numbers,

2x + 2 = -3x + 15 + 2

2x + 2 = -3x + 17

Subtract 2 from both sides,

2x + 2 = -3x + 17

2x + 2 - 2 = -3x + 17 - 2

2x  = -3x + 17 - 2

2x  = -3x + 15

Add 3x to both sides,

2x  = -3x + 15

2x + 3x  = -3x + 3x + 15

5x  = -3x + 3x + 15

5x  =  15

Divide both sides by the same factor (15),

5x  =  15

5x/5 = 15/5

x = 15/5

x = 3

Hence, the answer is 3.

4 0
2 years ago
58:27
Nina [5.8K]

Answer:

the answer is C, 18 inches width and 7 inches height

5 0
3 years ago
Read 2 more answers
Help? -Factoring polynomials
OLEGan [10]

Answer:

False

Step-by-step explanation:

If we have a factor then the remainder after the division will be zero.

8 0
3 years ago
Read 2 more answers
Given that (x+y+z)(xy+xz+yz)=18 and that x^2(y+z)+y^2(x+z)+z^2(x+y)=6 for real numbers x, y, and z, what is the value of xyz?
worty [1.4K]

Answer:

4

Step-by-step explanation:

(x+y+z)(xy+xz+yz)=18 Equation 1

x^2(y+z)+y^2(x+z)+z^2(x+y)=6 Equation 2

What is the value of xyz where each variable represents a real number?

Let's expand equation 1:

(x+y+z)(xy+xz+yz)=18

x(xy)+x(xz)+x(yz)+y(xy)+y(xz)+y(yz)+z(xy)+z(xz)+z(yz)=18

Simplify each term if can:

x^2y+x^2z+xyz+y^2x+xyz+y^2z+xyz+z^2x+z^2y=18

See if we can factor a little to get some of the left hand side of equation 2:

The first two terms have x^2 and if I factored x^2 from first two terms I would have x^2(y+z) which is the first term of left hand side of equation 2.

So let's see what happens if we gather the terms together that have the same variable squared together.

x^2y+x^2z+y^2x+y^2z+z^2y+z^2x+xyz+xyz+xyz=18

Factor the variable squared terms out of each binomial pairing:

x^2(y+z)+y^2(x+z)+z^2(y+x)+xyz+xyz+xyz=18

Replace the sum of those first three terms with what it equals which is 6 from the equation 2:

6+xyz+xyz+xyz=18

Combine like terms:

6+3xyz=18

Subtract 6 on both sides:

3xyz=12

Divide both sides by 3:

xyz=4

4 0
2 years ago
What is the domain of the following parabola? u-shaped graph that opens up with a vertex of 2, negative 3 All real numbers x ≥ 1
Shtirlitz [24]
The Answer is A, All real Numbers. Hope you do well!
4 0
2 years ago
Read 2 more answers
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