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Llana [10]
3 years ago
14

Solve the inequality 2x - 3

Mathematics
1 answer:
Crank3 years ago
6 0
That's not an inequality. 
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F(x) = 3x + 2; g(x) = x2 + 4; h(x) = 5x + 1<br><br> Find h(g(2))
tatuchka [14]

Answer:

Step-by-step explanation:

h(g(2)) = h( 2²)  = h(4) ; because g(x=2) = 2²

h(4) =  5*4+1 = 21

6 0
2 years ago
Mrs. barney earns $80 for tutoring for 2 hours. If she gets a 10% raise, how much does she now earn per hour.
olga_2 [115]

Answer:$88


Step-by-step explanation:

multiply 80 dollars by 10% then add that answer to 80 dollars and that gives you 88 dollars


8 0
3 years ago
WHAT IS X³-27 SIMPLIFIED
Eduardwww [97]

Answer:

<u>It</u><u> </u><u>is</u><u> </u><u>(</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>)</u><u>³</u><u> </u><u>-</u><u> </u><u>9</u><u>x</u><u>(</u><u>3</u><u> </u><u>-</u><u> </u><u>x</u><u>)</u>

Step-by-step explanation:

Express 27 in terms of cubes, 27 = 3³:

=  {x}^{3}  -  {3}^{3}

From trinomial expansion:

{(x - y)}^{3}  = (x - y)(x - y)(x - y) \\

open first two brackets to get a quadratic equation:

{(x - y)}^{3}  = ( {x}^{2}  - 2xy +  {y}^{2} )(x - y)

expand further:

{(x - y)}^{3}  =  {x}^{3}   - y {x}^{2}  - 2y {x}^{2}  + 2x {y}^{2}  + x {y}^{2}  -  {y}^{3}  \\  {(x - y)}^{3}  =  {x}^{3}  -  {y}^{3}  + 3x {y}^{2}  - 3y {x}^{2}  \\  {(x - y)}^{3}  =  {x}^{3}  -  {y}^{3}  + 3xy(y - x) \\  \\ { \boxed{( {x}^{3} -  {y}^{3} ) =  {(x - y)}^{3}   - 3xy(y - x)}}

take y to be 3, then substitute:

( {x}^{3}  - 3^3) =  {(x - 3)}^{3}  - 9x(3 - x)

5 0
3 years ago
HiCan you please state the Pythagorean TheoremVerbally and algebraically
Natalija [7]
Pythagorean Theorem<h2>Verbally:</h2>

Let's say a and b are the legs, and c is the hypotenuse. Then, algebraically, the theorem is,

a^2+b^2=c^2

7 0
1 year ago
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
3 years ago
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