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vladimir1956 [14]

3 years ago

15

How many diagonals can you draw from the vertex of an octagon

1 answer:

ludmilkaskok [199]3 years ago

8 0

From one vertex of an octagon you can draw 5 diagonals.

There are 8 vertices in an octagon, and we are choosing one as our starting vertex. There are then 7 vertices left to draw a line to, but 2 of the vertices are already connected to our main vertex (because they are connected along the side of the octagon). That leaves 5 vertices to draw a diagonal to from our original vertex.

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Match each number on the left with the correct description of the number on the right.

qwelly [4]

Answer:

im not sure the answer tho- just trying

4 0

2 years ago

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

andrew11 [14]

Differentiating both sides of

$x^2 + 2xy - y^2 + x = 20$

with respect to $x$ yields

$2x + 2y + 2x \dfrac{dy}{dx} - 2y \dfrac{dy}{dx} + 1 = 0 \\\\ \implies (2x-2y) \dfrac{dy}{dx} = -1 - 2x - 2y \\\\ \implies \dfrac{dy}{dx} = \dfrac{1 + 2x + 2y}{2(y-x)}$

At the point (3, 4) (so $x=3$ and $y=4$ ), the tangent line has slope

$\dfrac{dy}{dx} = \dfrac{1 + 2\times3 + 2\times4}{2(4-3)} = \dfrac{15}2$

Then the tangent line to (3, 4) has equation

$y - 4 = \dfrac{15}2 (x - 3) \implies \boxed{y = \dfrac{15}2 x - \dfrac{37}2}$

7 0

2 years ago

What does 22 divided by 3 plus 1 equal?

nordsb [41]

Answer:

8.33

Step-by-step explanation:

6 0

3 years ago

What are the Terms for 8x+6

densk [106]

Answer:Simplifying

8x + 6

Reorder the terms:

6 + 8x

Factor out the Greatest Common Factor (GCF), '2'.

2(3 + 4x)

Final result:

2(3 + 4x)

6 0

3 years ago

Someone help me out please

uysha [10]

1 step (B): raise both sides of the equation to the power of 2.

$(\sqrt{x+3}-\sqrt{2x-1})^2=(-2)^2,\\ (x+3)-2\sqrt{x+3}\cdot \sqrt{2x-1}+(2x-1)=4,\\ 3x+2-2\sqrt{x+3}\cdot \sqrt{2x-1}=4$ .

2 step (A): simplify to obtain the final radical term on one side of the equation.

$-2\sqrt{x+3}\cdot \sqrt{2x-1}=4-3x-2,\\ -2\sqrt{x+3}\cdot \sqrt{2x-1}=2-3x,\\ 2\sqrt{x+3}\cdot \sqrt{2x-1}=3x-2$ .

3 step (F): raise both sides of the equation to the power of 2 again.

$(2\sqrt{x+3}\cdot \sqrt{2x-1})^2=(3x-2)^2,\\ 4(x+3)(2x-1)=(3x-2)^2$ .

4 step (E): simplify to get a quadratic equation.

$4(2x^2-x+6x-3)=(3x)^2-2\cdot 3x\cdot 2+2^2,\\ 8x^2+20x-12=9x^2-12x+4,\\ x^2-32x+16=0$ .

5 step (D): use the quadratic formula to find the values of x.

$D=(-32)^2-4\cdot 16=1024-64=960, \\ \sqrt{D} =8\sqrt{5} ,\\ x_{1,2}=\dfrac{32\pm 8\sqrt{5}}{2} =16\pm 4\sqrt{5}$ .

6 step (C): apply the zero product rule.

$x^2-32x+16=(x-16-4\sqrt{5}) (x-16+4\sqrt{5}) ,\\ (x-16-4\sqrt{5}) (x-16+4\sqrt{5}) =0,\\ x_1=16+4\sqrt{5} ,x_2=16-4\sqrt{5}$ .

Additional 7 step: check these solutions, substituting into the initial equation.

3 0

3 years ago