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3 years ago

How is growth shown in a rule

Mathematics

1 answer:

alexandr1967 [171]3 years ago

4 0

Growth is shown when a number is increased within a rule. For example: x^2+5, 5 is how much it grows.

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Circle A is shown. Secant W Y intersects tangent Z Y at point Y outside of the circle. Secant W Y intersects circle A at point X

timama [110]

Circle A is missing, so i have attached it.

Answer:

∠XYZ = 35°

Step-by-step explanation:

We want to find the angle ∠XYZ in the image attached.

To solve that, we will use the formula in the theorem for angle formed by secants or tangents. Thus;

∠XYZ = ½(arc WZ - arc XZ)

From the image, arc WZ = 175° and arc XZ = 105°

Thus;

∠XYZ = ½(175 - 105)

∠XYZ = ½(70)

∠XYZ = 35°

5 0

3 years ago

Read 2 more answers

The square root of a number is between 8 and 9. Which of the following could be the value of that number? Select all that apply.

ser-zykov [4K]

What are the choices u have?<span />

7 0

3 years ago

Last year a construction worker had a gross income $ 29,700, of the which he contributed 7% to his 401(k) plain. If he got paid

AlladinOne [14]

Divide the gross income by 12 to find his monthly pay:

29,700 / 12 =$2,475 per month.

Multiply his monthly pay by 7%:

2,475 x 0.07 = $173.25 into his 401(k) per month.

The answer is A.

8 0

3 years ago

The price p of a pizza is $6.45 plus $0.90 for each of the t toppings on the pizza

Julli [10]

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3 years ago

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