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3 years ago

From the first step of the construction, drawing the arcs on both rays establishes that AB is

Mathematics

1 answer:

leonid [27]3 years ago

7 0

Answer: The answer is “Congruent to AC” !! When you are making an arc on both rays, they should measure/be exactly the same, and even more exact is you are using a compass

Step-by-step explanation:

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Use the properties of exponents to enter an equivalent expression with a single exponent.

Gelneren [198K]

$({10}^{2} ) {}^{4} = {10}^{2 \times 4} = {10}^{8}$

I hope this can help you ^_^

5 0

3 years ago

Jo's mother, Anne, is four times as old as Jo. In four years time, Anne will be three times as old as Jo. How old is Jo? (this i

pentagon [3]

Answer:

Jo's age = x = 8 years

Step-by-step explanation:

Let

Jo's age = x

Jo's mother, Anne's age = 4x

In four years time, Anne will be three times as old as Jo.

Jo's age =3( x + 4)

Jo's mother, Anne's age = 4x + 4

How old is Jo?

Equate Jo's age and his mother's age

3(x + 4) = 4x + 4

3x + 12 = 4x + 4

3x - 4x = 4 - 12

-x = - 8

Divide both sides by -1

x = 8

Therefore,

Jo's age = x = 8 years

8 0

3 years ago

Read 2 more answers

Given:

astraxan [27]

The answer is Vertical angles are equal.

5 0

3 years ago

Read 2 more answers

Graph y = tanx for -pi/4 ≤ x ≤ pi/4. What is the range?

Debora [2.8K]

Answer:

( -1, 1 )

Step-by-step explanation:

For f ( x ) = tan x ; Range = R (real no.)

Range in interval = ( -1, 1 )

3 0

3 years ago

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The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each

Fynjy0 [20]

Answer:

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Step-by-step explanation:

Given equation of ellipsoids,

$u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

The vector normal to the given equation of ellipsoid will be given by

$\vec{n}\ =\textrm{gradient of u}$

$=\bigtriangledown u$

$=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})$

$=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}$

$=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}$

Hence, the unit normal vector can be given by,

$\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}$

$=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}$

$=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Hence, the unit vector normal to each point of the given ellipsoid surface is

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

3 0

3 years ago