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Reil [10]
3 years ago
13

A polynomial function has -5+√3t as a root. Which of the following must also be a root of the function? a) -5-√3t b) -5+√3t c) 5

-√3t d) 5+√3t
Mathematics
1 answer:
juin [17]3 years ago
4 0

Answer:

a

Step-by-step explanation:

Radical roots occur as conjugate pairs

Given - 5 + \sqrt{3} t is a root , then

- 5 - \sqrt{3} t is also a root → a

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34=2m-8. Solve the equation
N76 [4]
The equation will be m=21
7 0
3 years ago
Read 2 more answers
Find the product (4x-5y)^2
sleet_krkn [62]
To find the product of (4x-5y)^2,
we can rewrite the problem as:
(4x-5y)(4x-5y) (two times because it is squared)

Now, time to use that old method we learned in middle school:
FOIL. (Firsts, Outers, Inners, and Lasts)

FOIL can help us greatly in this scenario.
Let's start by multiplying the 'Firsts' together:
4x * 4x = <em>16x^2</em>

Now, lets to the 'Outers':
4x * -5y = <em>-20xy</em> 

Next, we can multiply the 'Inners':
-5y * 4x = <em>-20xy</em>

Finally, let's do the 'Lasts':
-5y * -5y = <em>25y</em>^2

Now, we can take the products of these equations from FOIL and combine like terms. We have: 16x^2, -20xy, -20xy, and 25y^2.
-20xy and -20xy make -40xy.

The final equation (product of (4x-5y)^2) is:
16x^2 - 40xy + 25y^2

Hope I helped! If any of my math is wrong, please report and let me know!
Have a good one.
4 0
3 years ago
a BBQ restaurant grills 53 lb of chicken and one day. The restaurant does not close. How many pounds of chicken would the restau
nignag [31]

Answer: 19,345lb

Step-by-step explanation: 365(53)

4 0
2 years ago
Solve the expression x^2-100=0
enot [183]

Answer:

x = ± 10

Step-by-step explanation:

Given

x² - 100 = 0 ( add 100 to both sides )

x² = 100 ( take the square root of both sides )

x = ± \sqrt{100} ← note plus or minus, hence

x = ± 10

6 0
3 years ago
Simplify. Write your answer using whole numbers and variables.
Andre45 [30]

Answer:

q + 7

Step-by-step explanation:

(q² + 4q − 21) / (q − 3)

Factor the numerator:

(q + 7) (q − 3) / (q − 3)

Divide:

q + 7

5 0
3 years ago
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