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2 years ago

12

Solve 5x4/9. help pleeeeaaassee

2 answers:

Rus_ich [418]2 years ago

8 0

Answer:

$\sf \: 2 \frac{2}{9}$

Step-by-step explanation:

Q) 5 × (4/9) = ?

→ (5 × 4)/9

→ 20/9

$\rightarrow\sf \: 2 \frac{2}{9}$

Varvara68 [4.7K]2 years ago

8 0

Answer = 2 2/9
Explanation = 5 x 4/9
5 x 4 = 20
so we end up with 20/9
9 goes into 20 twice and were left with a reminder of 2
So the answer is 2 2/9

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Find the slope given the coordinate points.

romanna [79]

5 / 16

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3 years ago

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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:

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

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

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

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

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
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$\displaystyle m = -\frac{3}{4}$ ,
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The equation of this tangent line will thus be:

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

That simplifies to

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

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3 years ago

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Hi!

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3 years ago

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3 years ago

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UkoKoshka [18]

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