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3 years ago

Write the equation of the sinusoidal function shown.

Mathematics

1 answer:

Natalija [7]3 years ago

3 0

Answer: A

It looks at though the graph moved down a unit, so definitely a (-1) at the end of the function. If you move the graph up a unit, you will notice that the y = cos x format, therefore, it's not C or D.
The amplitude of the function is 1. So B and D are out because their amplitudes are 2.

<u>Therefore, the answer is A.</u>

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Matilda needs at least $112 to buy a new dress. She already has saved $40. She earns $9 an hour by babysitting. Write an inequal

Ymorist [56]

Answer:

Step-by-step explanation:

9*8 = 72 + 40 = 112

6 0

3 years ago

Is the function continuous at x = -17

Pavlova-9 [17]

For a function to be continuous at an x-value, say -17, you need to make sure two things line up:

The limit from the left equals the limit from the right.

$\lim_{x \to -17^{-}} f(x) = \lim_{x \to -17^{+}} f(x)$

This limit equals the functions value.

$\lim_{x \to -17} f(x) = f(-17)$

The left hand limit involves the first piece, f(x) = 20x + 1:

$\begin{aligned} \lim_{x \to -17^{-}} f(x) &= \lim_{x \to -17^{-}} (20x+1)\\[0.5em]&= 20(-17)+1\\[0.5em]&= -339\endaligned}$

The right hand limit invovles the second piece, f(x) = -10x^2:

$\begin{aligned} \lim_{x \to -17^{+}} f(x) &= \lim_{x \to -17^{+}} (-10x^2)\\[0.5em]&= -10\cdot (-17)^2\\[0.5em]&= -2890\endaligned}$

Since the two one-sided limits don't match, the function is not continuous at x=-17.

3 0

3 years ago

Consider the circle of radius 5 centered at (0, 0). Find an equation of the line tangent to the circle at the point (3, 4) in sl

Wittaler [7]

Answer:

$\displaystyle y= -\frac{3}{4} x + \frac{25}{4}$ .

Step-by-step explanation:

The equation of a circle of radius $5$ centered at $(0,0)$ is:

$x^{2} + y^{2} = 5^{2}$ .

$x^{2} + y^{2} = 25$ .

Differentiate implicitly with respect to $x$ to find the slope of tangents to this circle.

$\displaystyle \frac{d}{dx}[x^{2} + y^{2}] = \frac{d}{dx}[25]$

$\displaystyle \frac{d}{dx}(x^{2}) + \frac{d}{dx}(y^{2}) = 0$ .

Apply the power rule and the chain rule. Treat $y$ as a function of $x$ , $f(x)$ .

$\displaystyle \frac{d}{dx}(x^{2}) + \frac{d}{dx}(f(x))^{2} = 0$ .

$\displaystyle \frac{d}{dx}(2x) + \frac{d}{dx}(2f(x)\cdot f^{\prime}(x)) = 0$ .

That is:

$\displaystyle \frac{d}{dx}(2x) + \frac{d}{dx}\left(2y \cdot \frac{dy}{dx}\right) = 0$ .

Solve this equation for $\displaystyle \frac{dy}{dx}$ :

$\displaystyle \frac{dy}{dx} = -\frac{x}{y}$ .

The slope of the tangent to this circle at point $(3, 4)$ will thus equal

$\displaystyle \frac{dy}{dx} = -\frac{3}{4}$ .

Apply the slope-point of a line in a cartesian plane:

$y - y_0 = m(x - x_0)$ , where

$m$ is the gradient of this line, and
$(x_0, y_0)$ are the coordinates of a point on that line.

For the tangent line in this question:

$\displaystyle m = -\frac{3}{4}$ ,
$(x_0, y_0) = (3, 4)$ .

The equation of this tangent line will thus be:

$\displaystyle y - 4 = -\frac{3}{4} (x - 3)$ .

That simplifies to

$\displaystyle y= -\frac{3}{4} x + \frac{25}{4}$ .

3 0

3 years ago

If you follow the steps necessary to prove that square root 7 is irrational (use a proof by contradiction) to prove that square

Viefleur [7K]

Sorry, i wish i could help, but really the only thing i know is that 4 is 2 by 2 so using another equation to try to prove otherwise is irrational in itself

8 0

3 years ago

a cake divided into 10 equal slices.kim ate 3/5 of 1/2 of cake. what part of the whole cake did kim ate?

joja [24]

Answer:

3/10 of the whole cake

Step-by-step explanation:

When you divide a cake in halve you get 5 pieces on each side, if Kim ate 3/5 of 1/2 of the cake, then she ate 3 pieces, so she ate 3/10 of the cake, and other way to figure this out is multiplying the two fractions, 3/5 x 1/2 = 3/10

8 0

3 years ago