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SVETLANKA909090 [29]
2 years ago
8

Write a polynomial function of least degree with rational coefficients so that P(x)=0 has the given roots.

Mathematics
1 answer:
Fittoniya [83]2 years ago
3 0

Answer:

If P(x) has the roots -11 and 6, then its multipliers can be written as

Step-by-step explanation:

P(x)=(x+11)(x-6)

P(x)=x2-6x+11x-66

P(x) = x2+5x-66

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Solve the Equations
Vikentia [17]

Answer:

1) -2.13

2) 2.57

3) 33

Step-by-step explanation:

1)

3(8 + 5h) = -28  Distribute the 3

24 + 15h = -28  Subtract 24 from both sides of the equation

15h = -32  Divide both sides by 15 and round

h = - 2.13

2)

19 = 7(3n - 5)  Distribute the 7

19 = 21n - 35  Add 35 to both sides of the equation

54 = 21n  Divide both sides of the equation by 21

2.57 = n

3)

6s - 7s = -33  Combine the s's

-s = -33  Multiply both sides by -1

s = 33

5 0
2 years ago
Use the order of operations to simplify the given expression.<br> 4−6+2⋅7
andrezito [222]

Answer:

19

Step-by-step explanation:

PEMDAS

no parenthesis

no exponents

4-6+2 x 7

4-6+ 17

no division

4-6=2

2+17

19 is the answer

6 0
3 years ago
Read 2 more answers
Round 41.805 to the nearest whole number
aleksandr82 [10.1K]
The answer to your question is 42. The rule is 5 or more go up 1 . If the number is 4 les stay where it is. hope this help.

5 0
3 years ago
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Write an equation for the cubic polynomial function whose graph has zeroes at 2, 3, and 5.
TiliK225 [7]

we know that

For a polynomial, if x=a is a zero of the function, then (x−a) is a factor of the function. The term multiplicity, refers to the number of times that its associated factor appears in the polynomial.

So

In this problem

If the cubic polynomial function has zeroes at 2, 3, and 5

then

the factors are

(x-2)\\ (x-3)\\ (x-5)

Part a) Can any of the roots have multiplicity?

The answer is No

If a cubic polynomial function has three different zeroes

then

the multiplicity of each factor is one

For instance, the cubic polynomial function has the zeroes

x=2\\ x=3\\ x=5

each occurring once.

Part b) How can you find a function that has these roots?

To find the cubic polynomial function multiply the factors and equate to zero

so

(x-2)*(x-3)*(x-5)=0\\ (x^{2} -3x-2x+6)*(x-5)=0\\ (x^{2} -5x+6)*(x-5)=0\\ x^{3} -5x^{2} -5x^{2} +25x+6x-30=0\\ x^{3}-10x^{2} +31x-30=0

therefore

the answer Part b) is

the cubic polynomial function is equal to

x^{3}-10x^{2} +31x-30=0

7 0
3 years ago
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Eddie ended round one of a game show with 800 points. In round two, he scored -400 points
german

Answer:

400

Step-by-step explanation:

800-400=400

8 0
3 years ago
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