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Neko [114]
2 years ago
6

Please help please!!

Mathematics
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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Tatiana [17]

Answer:

\frac{2x}{x-1}

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2 years ago
Explain how finding 4 x 384 can help you find 4 * 5384 then find both products
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La ès pasanodáo mel lachillo
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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
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