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Margarita [4]
3 years ago
6

Y=(x^4)-(6^3)+(50x^2)+122x=65find the zeros

Mathematics
1 answer:
Jobisdone [24]3 years ago
7 0
Answer: I worked at the pizza place in Chucky cheese since i was little. I grew up and became manager. Then i sabotaged everyone there. I’m gonna cry right now . miss that job so much. I think I might….
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Multiply the given polynomials. Answer should be simplified and written in standard form.
vampirchik [111]

Answer: 14k²-99k+52.

Step-by-step explanation:

(2k-13)*(7k-4)=\\2k*(7k-4)-13*(7k-4)=\\14k^2-8k-91k+52=\\14k^2-99k+52.

8 0
2 years ago
Choose the correct simplification of (6x-5)(2x^2-3x-6)
bogdanovich [222]

Answer:

12x^3 -28x^2 -21x +30

Step-by-step explanation:

1) Distribute!

(6x-5)(2x^2-3x-6) = 12x^3 -18x^2 - 36x -10x^2 +15x+30

2) Combine like terms!

12x^3 -18x^2 - 36x -10x^2 +15x+30 = 12x^3 -28x^2 -21x +30

Answer is:

12x^3 -28x^2 -21x +30

Hope it helps!!

3 0
3 years ago
Can Somebody help me real quick?
Eddi Din [679]

Answer:The answer is (4 x -1)

Step-by-step explanation: How to solve this is tell the teacher I don’t know

8 0
3 years ago
What is x. Please show *all* the steps.
Zolol [24]

The equation 5/2 - x + x - 5/x + 2 + 3x + 8/x^2 - 4 = 0 is a quadratic equation

The value of x is 8 or 1

<h3>How to determine the value of x?</h3>

The equation is given as:

5/2 - x + x - 5/x + 2 + 3x + 8/x^2 - 4 = 0

Rewrite as:

-5/x - 2 + x - 5/x + 2 + 3x + 8/x^2 - 4 = 0

Take the  LCM

[-5(x + 2) + (x -5)(x- 2)]\[x^2 - 4 + [3x + 8]/[x^2 - 4] = 0

Expand

[-5x - 10 + x^2 - 7x + 10]/[x^2 - 4] + [3x + 8]/[x^2 - 4] = 0

Evaluate the like terms

[x^2 - 12x]/[x^2 - 4] + [3x + 8]/[x^2 - 4 = 0

Multiply through by x^2 - 4

x^2 - 12x+ 3x + 8 = 0

Evaluate the like terms

x^2 -9x + 8 = 0

Expand

x^2 -x - 8x + 8 = 0

Factorize

x(x -1) - 8(x - 1) = 0

Factor out x - 1

(x -8)(x - 1) = 0

Solve for x

x = 8 or x = 1

Hence, the value of x is 8 or 1

Read more about equations at:

brainly.com/question/2972832

8 0
2 years ago
Can u please help me on part C? The arcs in the photo at the right appear to be paths of stars rotating about the North Star. To
Natasha_Volkova [10]

no i would if i could but sadly i dont know the answer.

5 0
3 years ago
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