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Archy [21]
3 years ago
7

A system of equations is shown below:

Mathematics
1 answer:
JulijaS [17]3 years ago
6 0
X+y=3

Subtract x from both sides

y=-x+3

Substitute

2x--x+3=6
2x+x+3=6
3x+3=6

Subtract 3 from both sides

3x=3

Divide both sides by 3

x=1
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What is the domain and range? <br><br> {(-7,5), (-6,1), (-5,0), (-4,1)}
Monica [59]
The Domain is the first number of the pair
D= -7, -6, -5, -4
The Range is the second number of the pair
R= 5, 1, 0, when a number repeats like the number one you can put it once.


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Is it <br><br> A)6<br> B)9<br> C)16<br> D)18
joja [24]
18 because that’s the greatest common factor for all of them
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Calculate the discriminant to determine the number solutions. y = x ^2 + 3x - 10
Nataly_w [17]

1. The first step is to find the discriminant itself. Now, the discriminant of a quadratic equation in the form y = ax^2 + bx + c is given by:

Δ = b^2 - 4ac

Our equation is y = x^2 + 3x - 10. Thus, if we compare this with the general quadratic equation I outlined in the first line, we would find that a = 1, b = 3 and c = -10. It is easy to see this if we put the two equations right on top of one another:

y = ax^2 + bx + c

y = (1)x^2 + 3x - 10

Now that we know that a = 1, b = 3 and c = -10, we can substitute this into the formula for the discriminant we defined before:

Δ = b^2 - 4ac

Δ = (3)^2 - 4(1)(-10) (Substitute a = 1, b = 3 and c = -10)

Δ = 9 + 40 (-4*(-10) = 40)

Δ = 49 (Evaluate 9 + 40 = 49)

Thus, the discriminant is 49.

2. The question itself asks for the number and nature of the solutions so I will break down each of these in relation to the discriminant below, starting with how to figure out the number of solutions:

• There are no solutions if the discriminant is less than 0 (ie. it is negative).

If you are aware of the quadratic formula (x = (-b ± √(b^2 - 4ac) ) / 2a), then this will make sense since we are unable to evaluate √(b^2 - 4ac) if the discriminant is negative (since we cannot take the square root of a negative number) - this would mean that the quadratic equation has no solutions.

• There is one solution if the discriminant equals 0.

If you are again aware of the quadratic formula then this also makes sense since if √(b^2 - 4ac) = 0, then x = -b ± 0 / 2a = -b / 2a, which would result in only one solution for x.

• There are two solutions if the discriminant is more than 0 (ie. it is positive).

Again, you may apply this to the quadratic formula where if b^2 - 4ac is positive, there will be two distinct solutions for x:

-b + √(b^2 - 4ac) / 2a

-b - √(b^2 - 4ac) / 2a

Our discriminant is equal to 49; since this is more than 0, we know that we will have two solutions.

Now, given that a, b and c in y = ax^2 + bx + c are rational numbers, let us look at how to figure out the number and nature of the solutions:

• There are two rational solutions if the discriminant is more than 0 and is a perfect square (a perfect square is given by an integer squared, eg. 4, 9, 16, 25 are perfect squares given by 2^2, 3^2, 4^2, 5^2).

• There are two irrational solutions if the discriminant is more than 0 but is not a perfect square.

49 = 7^2, and is therefor a perfect square. Thus, the quadratic equation has two rational solutions (third answer).

~ To recap:

1. Finding the number of solutions.

If:

• Δ < 0: no solutions

• Δ = 0: one solution

• Δ > 0 = two solutions

2. Finding the number and nature of solutions.

Given that a, b and c are rational numbers for y = ax^2 + bx + c, then if:

• Δ < 0: no solutions

• Δ = 0: one rational solution

• Δ > 0 and is a perfect square: two rational solutions

• Δ > 0 and is not a perfect square: two irrational solutions

6 0
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PtichkaEL [24]
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Using the laws of indices simplify 625^3/8 ×5^1/2÷ 25​
Oxana [17]

Answer:

1

Step-by-step explanation:

\huge {625}^{ \frac{3}{8} }  \times  {5}^{ \frac{1}{2} }  \div 25 \\  \\ =  \huge {( {5}^{4})}^{ \frac{3}{8} }  \times  {5}^{ \frac{1}{2} }  \div  {5}^{2}  \\  \\  =   \huge{5}^{ \cancel4 \times  \frac{3}{ \cancel8 \:  \:  \red{ \bold 2}} }  \times  {5}^{ \frac{1}{2} }  \div  {5}^{2}  \\  \\   \huge=  {5}^{ \frac{3}{2} }  \times  {5}^{ \frac{1}{2} }  \div  {5}^{2}  \\  \\ \huge=  {5}^{ \frac{3}{2}  + \frac{1}{2}}  \div  {5}^{2}  \\ \\ \huge=  {5}^{ \frac{4}{2}  }  \div  {5}^{2}  \\ \\ \huge=  {5}^{ 2  }  \div  {5}^{2}  \\  \\   \huge=  {5}^{ 2  - 2 }  \\  \\   \huge=  {5}^{ 0 }  \\  \\   \huge=  1

5 0
2 years ago
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