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3 years ago

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Refer to the question below:

1 answer:

irakobra [83]3 years ago

7 0

∛(2x + 5) - 3 = 0
Add 3 to both sides
∛(2x + 5) = 3
Cube both sides
2x + 5 = 27
Subtract 5 from both sides
2x = 22
Divide both sides by 2
x=11
There's your answer. Have an awesome day! :)

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Anyone please???????

leva [86]

Answer:

$\boxed{\sf B)\: x=\sqrt{17}}$

Step-by-step explanation:

<u>Use the Pythagorean Theorem</u>

The equation for the Pythagorean Theorem is $\tt a^2+b^2=c^2$

* a and b are the lengths of the legs, and c is the hypotenuse.*

(To know which side is the hypotenuse look at longest side which is across the right angle).

a= x, b=8, c=9

$\sf a^2+b^2=c^2$

Let's Plug in the side lengths:

$\sf x^2+8^2=9^2$

Evaluate 8^2 and 9^2:

$\sf x^2+64=81$

Subtract 64 from both sides:

$\sf x^2+64-64=81-64$

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Take the square root of both sides:

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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:

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

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Rewrite as:

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

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

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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

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$2x =5$

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

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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:

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

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Answer:

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Step-by-step explanation:

-7r^2 = -329 . . . . given

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r = ±√47 . . . . square root both sides (square root property of equality)

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Answer:

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Step-by-step explanation:

I asked this same exact question, no one got the right answer, so I just guessed, and what do you know, I got the answer right! So here you go.

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