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2 years ago

How do I find the equation for a tangent line without derivatives?

Mathematics

1 answer:

Artist 52 [7]2 years ago

3 0

The equation of the tangent line is the remander when you divide by (x-a)^2

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3.5 × 4 brainliest...

arlik [135]

Answer:

Step-by-step explanation:

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2 years ago

Can someone help me with my home work and add explanation if u can ill give up to 120. points if there right, this is one of the

Varvara68 [4.7K]

Answer:

(5,300)

Step-by-step explanation:

its the only one that its Y axis cant be divided by 5 and get a equal/even number

sorry I cant explain it well, but hope I somewhat helped ^^

6 0

2 years ago

Read 2 more answers

I really need help I don’t understand how to do this

krek1111 [17]

Answer:

Step-by-step explanation:

Remark

The way this is worded, it sounds like it is a horizontal distance. However by the marking it looks like they want the hypotenuse. Life can be very confusing sometimes.

The angle of the triangle that you need to know is the one on the right. The line of sight line is parallel to the base of the triangle. So the angle on the right is 8 degrees by the Z theorem.

Solution

Sin(8) = 1817 / x Multiply both sides by x

x*sin(8) = 1817 Divide by sin(8)

x = 1817/sin(8)

sin(8) = 0.1392

x = 1817/0.1392

x = 13055.7

Note

Watch how you do this. I would put it into your calculator as

1817

sin(8)

If you use the rounded answer I gave for sin(8), you might get it wrong. This is a big distance and you should just use the exact number you get for sin(8).

8 0

2 years ago

Movies are on sale for 3/4 the original price. If the original price is $10, what is the sale price?

gtnhenbr [62]

3/4 of 10 would = $7.5 which would be your answer.

3 0

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

2 years ago