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3 years ago

7

Solve this quickly and CORRECTLY for brainliest

1 answer:

Bas_tet [7]3 years ago

4 0

You will have 40 servings

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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

Function f is shown on the graph below where two points are marked. If function f is horizontally compressed by a factor of 2, p

lilavasa [31]

Answer:

If you have

$f(x) = x^2$

The point (2,4) would be transformed to (1,1)

Step-by-step explanation:

If your compression is horizontal then the transformation you are making is the following

$g(x) = f(x/2)$

Therefore, if you have

$f(x) = x^2$

The point (2,4) would be transformed to (1,1)

5 0

3 years ago

Please help. I don't understand this.

klio [65]

Answer:

radius=6

Step-by-step explanation:

arc ACB=6xPi

Pi=3.14159

arc ACB=18.85 (rounded to 2 places)

ACB is a half circle so a full circle would be 18.85x2 or 37.7

circumference (whole circle)=2 x Pi x radius

37.7=2 x 3.14159 x radius

radius= 37.7/(2 x 3.14159)

radius=6

7 0

3 years ago

A student took a total of 4 tests over the course of 8 weeks. How many weeks of school will the student attend to take a total o

Julli [10]

Answer:

4/8 = 20/x

20 x 8 = 160

160 ÷ 4 = 40

the student will attend 40 weeks of school

3 0

3 years ago

Solve the exponential question . Leave your answer as a fraction

Doss [256]

Step-by-step explanation:

problem → 2^x = 4, solve for x

⇒2^x=4

⇒2^x=2^2

⇒x=2

7 0

3 years ago