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LuckyWell [14K]
2 years ago
12

Which of the following graphs is a polynomial function with x-intercepts of (-2, 0), (0, 0), and (3, 0)?

Mathematics
1 answer:
Sergio [31]2 years ago
8 0

Answer:

The intercepts of the third degree polynomial corresponds to the zeros of the equation

y = d*(x-a)*(x-b)(x-c)

Where a, b and c are the roots of the polynomial and d an adjustment coefficient.

y = d*(x+2)*(x)*(x-3)

Lets assume d = 1, and we get

y = (x+2)*(x)*(x-3) = x^3 - x^2 - 6x

We graph the equation in the attached file.

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If BE = 2x + 2, BD = 5x – 3, and AE = 4x – 6, what are the values of x and AC?
Keith_Richards [23]

Answer: x=7 and AC = 44 unuts.

Step-by-step explanation:

We know that the diagonals of a parallelogram bisect each other. (i)

Here in parallelogram ABCD , AC and Bd are diagonals intersecting at E.

BE = 2x + 2, BD = 5x – 3, and AE = 4x – 6

Using (i)

BE=\dfrac{BD}{2}\\\\2x+2=\dfrac{5x-3}{2}\\\\ 2(2x+2)=5x-3\\\\ 4x+4= 5x-3\\\\ 5x-4x=4+3\\\\ x= 7

Now , AE = 4(7)-6 = 28-6 = 22

AC =2 AE = 2 (22) =44 units.

Hence, x=7 and AC = 44 unuts.

3 0
3 years ago
The function g(x)=(x-2^2. The function f(x)=g(x) +3
OlgaM077 [116]
The parent function is:
 y = x ^ 2
 Applying the following function transformation we have:
 Horizontal translations:
 Suppose that h> 0
 To graph y = f (x-h), move the graph of h units to the right.
 We have then:
 g (x) = (x-2) ^ 2
 Then, we have the following function transformation:
 Vertical translations
 Suppose that k> 0
 To graph y = f (x) + k, move the graph of k units up.
 We have then that the original function is:
 g (x) = (x-2) ^ 2
 Applying the transformation we have
 f (x) = g (x) +3
 f (x) = (x-2) ^ 2 + 3
 Answer:
 
the function f(x)  moves horizontally 2 units rigth.
 
The function f (x) is shifted vertically 3 units up.
4 0
3 years ago
Differentiate 2sin(5x-3)
anygoal [31]
f(x)=2\sin(5x-3)\\
f'(x)=2\cos(5x-3)\cdot5=10\cos(5x-3)
3 0
3 years ago
What is this awnser? 2/5y=4
MA_775_DIABLO [31]

Answer:

\frac{2}{5}y=4\quad :\quad y=10

4 0
3 years ago
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jenyasd209 [6]
Parking lot A equation- $3.00 + $9.00x
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Part A- For 3 days parking lot B would be the better option- $28

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