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2 years ago

A baker is making rolls. He has 20 pounds of dough. Each roll is made from 1/8

Mathematics

1 answer:

larisa86 [58]2 years ago

6 0

ANSWER: he can make 160 rolls
EXPLANATION: each roll is 1/8 of a pound of dough, so 1 pound of the dough is 8 rolls

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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:

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

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

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

Please help me !!!

kati45 [8]

$23)~75r^2-48=3(25r^2-16)=3(5r-4)(5r+4)\\\\
24)~45x^2-5=5(9x^2-1)=5(3x-1)(3x+1)\\\\
25)~m^3n-mn^3=mn(m^2-n^2)=mn(m-n)(m+n)\\\\
26)~6a^3-24ab^2=6a(a^2-4b^2)=6a(a-2b)(a+2b)\\\\
27)~3x^2+15x-72=3(x^2+5x-24)=3(x-3)(x+8)\\\\
28)~2m^2-4m+2=2(m^2-2m+1)=2(m-1)^2\\\\
29)~x^3+9x^2-52x=x(x^2+9x-52)=x(x-4)(x+13)\\\\
30)~4y^3+4y^2-120y=4y(y^2+y-30)=4y(y-5)(y+6)$