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2 years ago

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Please help will mark brainliest

1 answer:

Wewaii [24]2 years ago

5 0

Answer:

you still there?

Step-by-step explanation:

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Find the distance, c, between (–3, 4) and (2, –3) on the coordinate plane. Round to the nearest tenth if necessary.

andriy [413]

Using the distance formula, the distance between (-3, 4) and (2, -3) is: 8.6 units.

<h3>What is the distance formula?</h3>

Distance formula for calculating the distance between two points on a coordinate plane is given as: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ .

Let,

(-3, 4) = (x1, y1)

(2, -3) = (x2, y2)

Plug in the values

$c = \sqrt{(2 - (-3))^2 + (-3 - 4)^2} \\\\c = \sqrt{74} \\\\$

c = 8.6 units.

Therefore, using the distance formula, the distance between (-3, 4) and (2, -3) is: 8.6 units.

Learn more about distance formula on:

brainly.com/question/661229

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1 year ago

What is 83/100 simplified?? PLEASE answer quick?

denpristay [2]

Answer:

Step-by-step explanation:

83/100 is already simplified. You can not make it smaller.

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2 years ago

Amber built a custom terrarium for her plants.What is the volume of the terrarium?

Alexus [3.1K]

Answer: 6600 cu in

Step-by-step explanation: (12*10*40=4800)+(10*10*18=1800)=6600

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2 years ago

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Verizon [17]

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2 years ago

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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

2 years ago