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3 years ago

Solve by using substitution. Express your answer as an ordered pair.

Mathematics

1 answer:

Furkat [3]3 years ago

3 0

Answer:

c. 2,4

Step-by-step explanation:

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Hi I need help with this problem. can anyone you tell me what 5+(-3)

Tomtit [17]

The answer for 5+(-3) is 2

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2 years ago

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Johnny can read 10 pages of a book in 15 minutes.

Svetach [21]

Answer:

Its C

Step-by-step explanation:

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2 years ago

susan designed a circular pool with a diameter of 24 meters. what is the area of the bottom of the pool?

Gennadij [26K]

Diameter = 24 m

radius = diameter/2 = 24 m / 2 = 12 m

area = pi * r^2 = 3.14 * (12 m)^2 = 3.14 * 144 m^2 = 452.16 m^2

7 0

2 years ago

Use the definition of the derivative to differentiate v=4/2 pie r^3

nata0808 [166]

I suspect 4/2 should actually be 4/3, since 4/2 = 2, while 4/3 would make V the volume of a sphere with radius r. But I'll stick with what's given:

$\displaystyle \frac{dV}{dr} = \lim_{h\to0} \frac{2\pi(r+h)^3-2\pi r^3}{h}$

$\displaystyle \frac{dV}{dr} = 2\pi \lim_{h\to0} \frac{(r^3+3r^2h+3rh^2+h^3)- r^3}{h}$

$\displaystyle \frac{dV}{dr} = 2\pi \lim_{h\to0} \frac{3r^2h+3rh^2+h^3}{h}$

$\displaystyle \frac{dV}{dr} = 2\pi \lim_{h\to0} (3r^2+3rh+h^2)$

$\displaystyle \frac{dV}{dr} = 2\pi \cdot 3r^2 = \boxed{6\pi r^2}$

In Mathematica, you can check this result via

D[4/2*Pi*r^3, r]

3 0

2 years ago

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$

Then

$y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2$

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

8 0

2 years ago