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3 years ago

14

All radical expressions that are equivalent to 3 2/5

1 answer:

SCORPION-xisa [38]3 years ago

4 0

Answer:

15

Step-by-step explanation:

√15 (3√5)^2 3√5^2 3√10

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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

Whats -2(x-3)=5(2x+3)

puteri [66]

-2(x - 3) = 5(2x + 3)

-2x + 6 = 10x + 15

6 = 10x + 15 + 2

6 = 12x + 15

6 - 15 = 12x

-9 = 12x

-9/12 = x

-3/4 = x or in decimal form; -0.75

hope this helps, God bless!

8 0

2 years ago

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