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2 years ago

6

Please help please!!

2 answers:

vodomira [7]2 years ago

7 0

The answer is 15 which is letter D

Darina [25.2K]2 years ago

5 0

The answer is D. Which is 15

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Simplify fully<br> <img src="https://tex.z-dn.net/?f=%5Cfrac%7B4x%5E%7B2%7D%20%2B4x%7D%7B2x%5E%7B2%7D%20-2%7D" id="TexFormula1"

Tatiana [17]

Answer:

$\frac{2x}{x-1}$

5 0

2 years ago

Explain how finding 4 x 384 can help you find 4 * 5384 then find both products

Lena [83]

La ès pasanodáo mel lachillo

6 0

3 years ago

The point (3, -4) is on the terminal side of an angle . What is cos ?

elena55 [62]

This is in the fourth quadrant where the cosine is positive
By pythagoras the hypotenuse is 5

so the cosine is 3/5 B

8 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

Domain and range of linear and quadratic functions worksheet

Vlad [161]

If the value of a is negative, then the range will be (-∞, k) and if the value of the a is positive then the range will be (k, ∞).

<h3>What is a quadratic equation?</h3>

It's a polynomial with a worth of nothing.

There exist polynomials of variable power 2, 1, and 0 terms.

A quadratic condition is a condition with one explanation where the degree of the equation is 2.

Domain and range of linear and quadratic functions

Let the linear equation be y = mx + c.

Then the domain and the range of the linear function are always real.

Let the quadratic equation will be in vertex form.

y = a(x - h)² + k

Then the domain of the quadratic function will be real.

If the value of a is negative, then the range will be (-∞, k) and if the value of the a is positive then the range will be (k, ∞).

More about the quadratic equation link is given below.

brainly.com/question/2263981

#SPJ4

7 0

1 year ago