Ask question

Ask question

All categories

3 years ago

9

Hi i need help on segment relationships in circles

1 answer:

exis [7]3 years ago

3 0

Just ask goku mannnnn

You might be interested in

What is the equation of the graphed line written in standard form?

Dmitry_Shevchenko [17]

Answer:

A

Step-by-step explanation:

4 0

3 years ago

This is what i need help on

Scilla [17]

G2=g1*2
g3=g1*2²
196=x*4
/4 /4
49

7 0

3 years ago

Something added by -8 equals 20

LiRa [457]

Answer:

28 is added by -8 equals 20

Step-by-step explanation:

The equation to calculate something added by -8 equals 20 is as follows:

x-8 = 20

x = 20+8

x = 28

28 is added by -8 equals 20

8 0

3 years ago

Read 2 more answers

Find the point (,) on the curve =8 that is closest to the point (3,0). [To do this, first find the distance function between (,)

ELEN [110]

Question:

Find the point (,) on the curve $y = \sqrt x$ that is closest to the point (3,0).

[To do this, first find the distance function between (,) and (3,0) and minimize it.]

Answer:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

Step-by-step explanation:

$y = \sqrt x$ can be represented as: $(x,y)$

Substitute $\sqrt x$ for $y$

$(x,y) = (x,\sqrt x)$

So, next:

Calculate the distance between $(x,\sqrt x)$ and $(3,0)$

Distance is calculated as:

$d = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2}$

So:

$d = \sqrt{(x-3)^2 + (\sqrt x - 0)^2}$

$d = \sqrt{(x-3)^2 + (\sqrt x)^2}$

Evaluate all exponents

$d = \sqrt{x^2 - 6x +9 + x}$

Rewrite as:

$d = \sqrt{x^2 + x- 6x +9 }$

$d = \sqrt{x^2 - 5x +9 }$

Differentiate using chain rule:

Let

$u = x^2 - 5x +9$

$\frac{du}{dx} = 2x - 5$

So:

$d = \sqrt u$

$d = u^\frac{1}{2}$

$\frac{dd}{du} = \frac{1}{2}u^{-\frac{1}{2}}$

Chain Rule:

$d' = \frac{du}{dx} * \frac{dd}{du}$

$d' = (2x-5) * \frac{1}{2}u^{-\frac{1}{2}}$

$d' = (2x - 5) * \frac{1}{2u^{\frac{1}{2}}}$

$d' = \frac{2x - 5}{2\sqrt u}$

Substitute: $u = x^2 - 5x +9$

$d' = \frac{2x - 5}{2\sqrt{x^2 - 5x + 9}}$

Next, is to minimize (by equating d' to 0)

$\frac{2x - 5}{2\sqrt{x^2 - 5x + 9}} = 0$

Cross Multiply

$2x - 5 = 0$

Solve for x

$2x =5$

$x = \frac{5}{2}$

Substitute $x = \frac{5}{2}$ in $y = \sqrt x$

$y = \sqrt{\frac{5}{2}}$

Split

$y = \frac{\sqrt 5}{\sqrt 2}$

Rationalize

$y = \frac{\sqrt 5}{\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$y = \frac{\sqrt {10}}{\sqrt 4}$

$y = \frac{\sqrt {10}}{2}$

Hence:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

3 0

3 years ago

HELP HELP HELP<br> Quick algebra 1 questions

beks73 [17]

<u>Hello:</u>

<u />

<u>Okay, this problem should be taken in a series of steps</u>:

<u>What is the slope's formula?</u>

$\rm\hookrightarrow \frac{y_2-y_1}{x_2-x_1}$

(x₁,y₁) -- first point
(x₂,y₂) -- second point

<u>Okay, now let's use this knowledge</u>

10. (5,-19) and (-5,21)

$Slope = \frac{21--19}{-5-5} =\frac{40}{-10}=-4$

11. (12,1) and (12, -1)

$Slope = \frac{-1-1}{12-12} =\frac{-2}{0}$

12. (8,7) and (4,7)

$Slope = \frac{7-7}{4-7} =0$

<u>Answer:</u>

Question 10: -4
Question 11: undefined
Question 12: 0

Hopefully that helps!

5 0

1 year ago