Jeff makes t-shirts. The shaded part below represents the fraction of a yard of fabric he uses for each t-shirt.

Find the area of a square with side a when a is equal to the following values. Are the area of a square and the length of its si

Xelga [282]

Answer:

The area of the square is 4 square cm. The area of the square and the length of its side are directional proportional quantities.

Step-by-step explanation:

A square has equal sides. We have to do length times height to get the area. 2*2 is 4. The area is 4 cm. And since this value needs to be multiplied by itself to get the area, it IS proportional.

3 0

3 years ago

Read 2 more answers

Use the inverse function-inverse cofunction

vivado [14]

Answer:

$arccot(x)'=-\frac{1}{1+x^2}$

Step-by-step explanation:

Use implicit differentiation:

$y=arccot(x)$

$cot(y)=x$

$\frac{1}{tan(y)}=x$

$\frac{cos(y)}{sin(y)}=x$

$\frac{dy}{dx}(\frac{cos(y)}{sin(y)})=\frac{dy}{dx}(x)$

$\frac{sin(y)(-sin(y))-cos(y)cos(y)}{sin^2(y)}*\frac{dy}{dx}=1$ (Quotient Rule: $\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$ )

$\frac{-sin^2(y)-cos^2(y)}{sin^2(y)}*\frac{dy}{dx}=1$

$\frac{-(sin^2(y)+cos^2(y))}{sin^2(y)}*\frac{dy}{dx}=1$

$\frac{-1}{sin^2(y)}*\frac{dy}{dx}=1$

$-csc^2(y)\frac{dy}{dx}=1$

$\frac{dy}{dx}=\frac{1}{-csc^2(y)}$

$\frac{dy}{dx}=-sin^2(y)$

$\frac{dy}{dx}=-sin^2(arccot(x))$

See the attached picture to understand how to evaluate the mixed composition of trig functions using a right triangle.

Therefore, the derivative of arccot(x) is $-\frac{1}{\sqrt{1+x^2}}$ .

4 0

2 years ago

Please answer What is the turning point of the graph of f(x) = |x + 4| ? (0, –4) (–4, 0) (4, 0) (0, 4)

saul85 [17]

X-intercept: (-4,0)
y-intercept: (0.4)

6 0

3 years ago

How many different ways can the first 12 letters of the alphabet be arranged?

zhannawk [14.2K]

There are 479001600 different ways the first 12 letters of the alphabet can be arranged.

3 0

3 years ago

Which graph represents the function f (x) = StartFraction 2 Over x minus 1 EndFraction + 4?

Pepsi [2]

Answer:

On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4

Step-by-step explanation:

The given function is presented as follows;

$f(x) = \dfrac{2}{x - 1} + 4$

From the given function, we have;

When x = 1, the denominator of the fraction, $\dfrac{2}{x - 1}$ , which is (x - 1) = 0, and the function becomes, $\dfrac{2}{1 - 1} + 4 = \dfrac{2}{0} + 4 = \infty + 4 = \infty$ therefore, the function in undefined at x = 1, and the line x = 1 is a vertical asymptote

Also we have that in the given function, as x increases, the fraction $\dfrac{2}{x - 1}$ tends to 0, therefore as x increases, we have;

$\lim_ {x \to \infty} \dfrac{2}{(x - 1)} \to 0, and \ \dfrac{2}{(x - 1)} + 4 \to 4$

Therefore, as x increases, f(x) → 4, and 4 is a horizontal asymptote of the function, forming a curve that opens up and to the right in quadrant 1

When -∞ < x < 1, we also have that as x becomes more negative, f(x) → 4. When x = 0, $\dfrac{2}{0 - 1} + 4 = 2$ . When x approaches 1 from the left, f(x) tends to -∞, forming a curve that opens down and to the left

Therefore, the correct option is on a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4.

5 0

2 years ago