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3 years ago

6

How many turning points can a polynomial with a degree of 7 have?

1 answer:

Gre4nikov [31]3 years ago

6 0

The best answer to this question is A because the graph of a polynomial can have up to 1 less turning point than its highest degree. For example, a 10th degree polynomial can have 9 or less turning points.

You might be interested in

What is -a^-2 if a=-5

Leviafan [203]

$-a^{-2}=-(-5)^{-2}=-\left(\dfrac{1}{-5}\right)^2=-\dfrac{1}{25}$

3 0

3 years ago

i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

3 0

3 years ago

A circle with a radius of 12 inches has an arc that measures 8 pi inches. Find the measure of the central angle determined by th

ivanzaharov [21]

We know that

[length of a circle]=2*pi*r
r=12 in
[length of a circle]=2*pi*12--------------> 24pi in

if 360° (full circle)--------------------> has a length of 24pi in
X-------------------------------------------> 8pi

X=8pi*360/24pi-----------> 120°

the answer is 120°

3 0

3 years ago

Nvm i got it lol sorry!

Dima020 [189]

Lol that’s good to hear I’m glad

4 0

2 years ago

63 POINT!!!!! GUYS, I NEED HELP WITH PLSSSSS NO CHEAT THANK YOU!!

nikdorinn [45]

Answer:

236

Step-by-step explanation:

We assume the cost is a linear function of the number of lessons. That means the slope is the same everywhere.

Let (lessons, cost) = (x, y). We have (4, 84), (8, 160), (12, y3).

(y2 -y1)/(x2 -x1) = (y3 -y2)/(x3 -x2)

(160 -84)/(8 -4) = (y3 -160)/(12 -8)

76 = y3 -160

y3 = 160 +76 = 236

The cost will be $236 for 12 lessons.

__

The cost function is ...

y = 19x +8 . . . . for x lessons

6 0

2 years ago