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KengaRu [80]
2 years ago
6

What is the polynomial answer for (-3+3n5+6n3)+(7n5+5n3-3)​

Mathematics
2 answers:
Gennadij [26K]2 years ago
7 0

10n^5+11n^3-6 is the answer

hope this helps

Serjik [45]2 years ago
5 0

Answer:

  10n^5 +11n^3 -6

Step-by-step explanation:

Add like terms:

  (-3+3n^5+6n^3)+(7n^5+5n^3-3)

  = n^5(3 +7) +n^3(6 +5) +(-3 -3)

  = 10n^5 +11n^3 -6

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Identify the zeros of the function f(x) = 2x^2 − 4x + 5 using the Quadratic Formula
Nostrana [21]

For this case we have that by definition, the roots, or also called zeros, of the quadratic function are those values of x for which the expression is 0.

Then, we must find the roots of:

2x ^ 2-4x + 5 = 0

Where:

a = 2\\b = -4\\c = 5

We have to:x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}

Substituting we have:

x = \frac {- (- 4) \pm \sqrt {(- 4) ^ 2-4 (2) (5)}} {2 (2)}\\x = \frac {4 \pm \sqrt {16-40}} {4}\\x = \frac {4 \pm \sqrt {-24}} {4}

By definition we have to:

i ^ 2 = -1

So:

x = \frac {4 \pm \sqrt {24i ^ 2}} {4}\\x = \frac {4 \pm i \sqrt {24}} {4}\\x = \frac {4 \pm i \sqrt {2 ^ 2 * 6}} {4}\\x = \frac {4 \pm 2i \sqrt {6}} {4}\\x = \frac {2 \pm i \sqrt {6}} {2}

Thus, we have two roots:

x_ {1} = \frac {2 + i \sqrt {6}} {2}\\x_ {2} = \frac {2-i \sqrt {6}} {2}

Answer:

x_ {1} = \frac {2 + i \sqrt {6}} {2}\\x_ {2} = \frac {2-i \sqrt {6}} {2}

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3 years ago
Kevin has had a checking account for a month. As he reconciles his account, Kevin notices that the dates on his check register d
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Find the slope of the line passing through (-9,3), (-3,-5)
egoroff_w [7]

Answer:

-4/3

Step-by-step explanation:

Slope: (y2-y1)/(x2-x1)

(-5-3)/(-3+9) = -8/6 = -4/3

The slope is -4/3

6 0
3 years ago
Please help 2х -3y =18
siniylev [52]

Answer:

x = 3(6 + y)/2

Step-by-step explanation:

Solving for x

Add 3y to both sides.

2x = 18 + 3y

Divide both sides by 2.

x = 18 + 3y/2

Factor out the common term 3.

x = 3(6 + y)/2

3 0
2 years ago
What is the length of the tangent line in the picture?
S_A_V [24]

Answer:

x = -2

Step-by-step explanation:

The tangent line has length "x + 8"

The secant line has length "x + 6 + 5", where

5 is the inner part

x + 6 is the outer part

Now,

the secant-tangent theorem tells us that square of the tangent line is equal to the outer segment of secant line multiplied by length of whole secant line.

So, we can say:

(x+8)^2 = (x+6)(x+6+5)

We can solve for x shown below:

(x+8)^2 = (x+6)(x+6+5)\\(x+8)^2=(x+6)(x+11)\\x^2+16x+64=x^2+17x+66\\17x-16x=64-66\\x=-2

The value of x is -2

4 0
3 years ago
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