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2 years ago

-3x° (2x7 - 4x + 1) multiply through polynomial

Mathematics

1 answer:

valina [46]2 years ago

7 0

Answer:

-6x⁷ + 12x − 3

Step-by-step explanation:

<u><em>Method 1</em></u> :

$-3x^0\ (2x^7-4x+1)$

$= (-3x^0)\times 2x^7-(-3x^0)\times4x+(-3x^0)\times1$

$= (-3x^0)\times 2x^7+(3x^0)\times4x+(-3x^0)\times1$

$=-6x^{0+7}+12x^{0+1}-3x^0$

$=-6x^{7}+12x^{1}-3x^0$

$=-6x^{7}+12x-3$

<u><em>Method 2</em></u> :

Note : x⁰ = 1

$-3x^0\ (2x^7-4x+1)$

$=-3\ (2x^7-4x+1)$

$=-3\times2x^7-(-3)\times4x+(-3)\times1$

$=-6x^7+12x-3$

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The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of

defon

The abscissa of the ordered pair, that is the x-coordinate, is equal to 1 and the ordinate, the y-coordinate, is equal to -1. In the cartesian plane, this point lies in the fourth (IV) quadrant. The standard position of the angle is that which has one of its side is in the x-axis.

Solve for the hypotenuse of the right triangle formed.
h = sqrt((-1)² + (1)²) = √2
Below items show the calculation for each of the trigonometric functions.

sin θ = opposite/hypotenuse = y/h = (-1)/(√2) = -√2/2
cos θ = adjacent/hypotenuse = x/h = (1)/√2 = √2/2
tan θ = opposite/adjacent = y/x = -1/1 = -1

7 0

3 years ago

HELP PRECALC I DO NOT UNDERSTAND AT ALLLLL!!!!!!!!!!!!!!!!!!!!!!

Arlecino [84]

Answer:

φ ≈ 1.19029 radians (≈ 68.2°)

Step-by-step explanation:

There are simple formulas for A and φ in this conversion, but it can be instructive to see how they are derived.

We want to compare ...

y(t) = Asin(ωt +φ)

y(t) = Psin(ωt) +Qcos(ωt)

Using trig identities to expand the first equation, we have ...

y(t) = Asin(ωt)cos(φ) +Acos(ωt)sin(φ)

Matching coefficients with the second equation, we have ...

P = Acos(φ)

Q = Asin(φ)

The ratio of these eliminates A and gives a relation for φ:

Q/P = sin(φ)/cos(φ)

Q/P = tan(φ)

φ = arctan(Q/P) . . . . taking quadrant into account

We can also use our equations for P and Q to find A:

P² +Q² = (Acos(φ))² +(Asin(φ))² = A²(cos(φ)² +sin(φ)²) = A²

A = √(P² +Q²)

_____

Here, we want φ.

φ = arctan(Q/P) = arctan(5/2)

φ ≈ 1.19029 . . . radians

8 0

4 years ago

I need the domain range and function. With an explanation

erastovalidia [21]

<h3>Answers:</h3>

Problem 1

Domain = $-3 < x \le 3$ , interval notation (-3, 3]
Range = $-3 \le y < 3$ , interval notation [-3, 3)
Is it a function? Yes

Problem 2

Domain = $x \ge -2$ , interval notation $[-2, \infty)$
Range = All real numbers, interval notation $(-\infty, \infty)$
Is it a function? No

Problem 3

Domain = $-4 \le x < 3$ , interval notation [-4, 3)
Range = $-4 < y \le 3$ , interval notation (-4, 3]
Is it a function? Yes

Problem 4

Domain = All real numbers, interval notation $(-\infty, \infty)$
Range = $y \le 4$ , interval notation $(-\infty, 4]$
Is it a function? Yes

==================================================

Explanations:

The left most point is when x = -3, and we are not including this value due to the open hole. The other endpoint is included because it is a filled in circle. The domain is therefore $-3 < x \le 3$ which in interval notation is (-3, 3]. We have the curved parenthesis meaning "exclude endpoint" and the square bracket says "include endpoint". The range is a similar story but we're looking at the smallest and largest y values. Though be careful about which endpoint is open/closed. We have a function because it passes the vertical line test.
The smallest x value is x = -2. There is no largest x value because the arrows say to go on forever to the right. We can say the domain is $x \ge -2$ which in interval notation is $[-2, \infty)$ . The range is $(-\infty, \infty)$ to indicate "all real numbers". This graph fails the vertical line test, so it is not a function. The vertical line test is where we check to see if we can pass a vertical line through more than one point on the curve. In this case, such a thing is possible which is why it fails the test.
This is the same idea as problem 1, though note the endpoints are flipped in terms of which has an open circle and which doesn't. It is not possible to draw a single vertical line to have it pass through more than one point on the curve, so it passes the vertical line test and we have a function.
This is a function because it passes the vertical line test. The domain is the set of all real numbers due to the arrows in both directions. Any x value is a possible input. The range is $y \le 4$ which is the same as saying $(-\infty, 4]$ in interval notation. This is because y = 4 is the largest y value possible. There is no smallest y value due to the arrows.