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olga55 [171]
2 years ago
15

How do you know if a function is quadratic??

Mathematics
2 answers:
Anestetic [448]2 years ago
7 0

Answer:

Usually it's in standard form: ax²+bx+c

or vertex form: y=a(x-h)²+k

If you put the graph into a graphing calculator it's going to look like a hill/depression.

Step-by-step explanation:

Standard form ex: 3x²+2x+1

Vertex form ex: y=4(x-1)²-2

butalik [34]2 years ago
6 0
F(x) = y = ax^2 + bx + c is the equation, where a, b, and c are numbers and a is not equal to zero. If it was zero, then the equation would be linear and not quadratic. For example, if you have f(x) = x +9 + 4x^2, you would rewrite the function with the largest exponent first, so you would have f(x) = 4x^2 + x + 9.
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Square route of fifty six
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Hope this helps!

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2 years ago
The quotient of two numbers is 21. Their difference is 60.What are the two numbers?
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The numbers are 63 and 3
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3 years ago
Find the real numbers x and y if -3+ix^2y and x^2+y+4i are conjugate of each other. Pls solve with the steps
Firdavs [7]
ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work

EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same

For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y

For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4

Therefore, for the two expressions to be conjugates, we must satisfy the two conditions. 

Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the 

   x²y = -4 ... (I)

Condition 2: Real parts are the same

   x² + y = -3 ... (II)

We have a system of equations since both conditions must be satisfied

   x²y = -4 ... (I)
   x² + y = -3 ... (II)

We can rearrange equation (II) so that we have

   y = -3 - x² ... (II)

Substituting into equation (I)

   x²y = -4 ... (I)
   x²(-3 - x²) = -4
   -3x² - x⁴ = -4
   x⁴ + 3x² - 4 = 0
   (x² + 4)(x² - 1) = 0
   (x² + 4)(x-1)(x+1) = 0

Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.

Solve for y:

   y = -3 - x² ... (II)
   y = -3 - (±1)²
   y = -3 - 1
   y = -4

So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:

   -3 + ix²y 
   = -3 + i(±1)²(-4)
   = -3 - 4i

   x² + y + 4i
   = (±1)² - 4 + 4i
   = 1 - 4 + 4i
   = -3 + 4i

They result in conjugates
4 0
3 years ago
Read 2 more answers
Mark has $18.00. What is the greatest number of days he can rent a bicycle? 2d=18 where d represents the number of days of the r
aksik [14]

Answer:

9

Step-by-step explanation:

9x2 is 18, so he can rent a bike for nine days if the rent is $2

6 0
2 years ago
She mowed 10 lawns for $90. How much did she make per lawn?
Evgen [1.6K]

Answer:

9$ per lawn

Step-by-step explanation:

8 0
2 years ago
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