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3 years ago

Write the polynomial in standard form. Then name the polynomial based on it's degree and number of terms

1 answer:

3 0

First simplify the polynomial by combining terms with the same variable. This is shown below.
5 - 12x + (13x² - 4x²)
5 - 12x + 9x²
Next, arrange the terms in such a way that the exponent of the variable x is decreasing,
9x² - 12x + 5
This is a quadratic trinomial.

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(b) (2x-1)(3x+5)-(2x - 1)(x+4)

QveST [7]

Answer:

4 x^2 - 1, this would be the answer to your question.

8 0

4 years ago

Read 2 more answers

In the Olympics, how many possible ways can 12 biathletes be awarded the gold, silver and bronze medals?

Leya [2.2K]

Well if you look closley

you have: gold sivler bronze metals (thats 3 medals) divide that by 12 and you get 4 so each could get four medals each

3 0

3 years ago

Q 2 PLEASE HELP ME FIGURE THIS OUT

suter [353]

Answer: IV, positive, $\frac{\pi} {6}$ , - sec $\frac{\pi} {6}$ , $\frac{2\sqrt{3}}{3}$

Step-by-step explanation:

a) Look at the Unit Circle to see that $\frac{11\pi} {6}$ = 330°, which is located in Quadrant IV.

b) The coordinate (cos θ, sin θ) for $\frac{11\pi} {6}$ is: $(\frac{\sqrt{3}} {2},\frac{-1}{2})$

sec = $\frac{1}{cos}$ = $\frac{2}{\sqrt{3}}$ which is positive

c) Since the given angle is in Quadrant IV, which is closest to the x-axis at 360° = 2π, the reference angle can be found by subtracting the given angle $\frac{11\pi} {6}$ from 2π: $\frac{12\pi} {6}$ - $\frac{11\pi} {6}$ = $\frac{\pi} {6}$

d) the reference angle is below the x-axis so the given angle is equal to the negative of the reference angle: - sec $\frac{\pi} {6}$ .

e) sec $\frac{11\pi} {6}$ = $\frac{2}{\sqrt{3}}$ = $\frac{2}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}}$ = $\frac{2\sqrt{3}}{3}$

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Answer: $\frac{18\pi}{11}$ , IV, $\frac{4\pi} {11}$

Step-by-step explanation:

2π is one rotation. 2π = $\frac{22\pi}{11}$

$\frac{-26\pi}{11}$ + $\frac{22\pi}{11}$ = $\frac{-4\pi}{11}$

$\frac{-4\pi}{11}$ + $\frac{22\pi}{11}$ = $\frac{18\pi}{11}$

Convert the radians into degrees to see which Quadrant it is in by setting up the proportion and cross multiplying:

$\frac{\pi}{180}$ = $\frac{18\pi}{11x}$

π(11x) = (180)18π

x = $\frac{180(18\pi}{11\pi}$

x = 295° which lies in Quadrant IV

Since the given angle is in Quadrant IV, which is closest to the x-axis at 360° = 2π, the reference angle can be found by subtracting the angle of least nonegative value $\frac{18\pi} {11}$ from 2π: $\frac{22\pi} {11}$ - $\frac{18\pi} {11}$ = $\frac{4\pi} {11}$

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Answer: $\frac{5\pi}{3}$ , IV, $\frac{4\pi} {11}$ , $\frac{\pi} {3}$

Step-by-step explanation:

2π is one rotation. 2π = $\frac{6\pi}{3}$

$\frac{-13\pi}{3}$ + $\frac{6\pi}{3}$ = $\frac{-7\pi}{3}$

$\frac{-7\pi}{3}$ + $\frac{6\pi}{3}$ = $\frac{-\pi}{3}$

$\frac{-\pi}{3}$ + $\frac{6\pi}{3}$ = $\frac{5\pi}{3}$

This is on the Unit Circle at 300°, which is located in Quadrant IV

Since the given angle is in Quadrant IV, which is closest to the x-axis at 360° = 2π, the reference angle can be found by subtracting the angle of least nonegative value $\frac{5\pi} {3}$ from 2π: $\frac{6\pi} {3}$ - $\frac{5\pi} {3}$ = $\frac{\pi} {3}$