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2 years ago

Need help fast please with these 5 questions

Mathematics

1 answer:

Natali [406]2 years ago

3 0

Answer:

16.....................................

6⁹*6ⁿ = 6¹²
6⁹⁺ⁿ = ¹²
9 + n = 12
n = 3

17.....................................

7⁸/7ⁿ = 7³
7⁸⁻ⁿ = 7³
8 - n = 3
n = 5

18.....................................

15ⁿ = 1
15ⁿ = 15⁰
n = 0

19.....................................

1/10⁵ = 10ⁿ
10⁻⁵ = 10ⁿ
-5 = n
n = -5

20.....................................

(8²)ⁿ = 8¹²
8²ⁿ = 8¹²
2n = 12
n = 6

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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:

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

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Given that the measure of X is 57° and the measure of y is 51° find the measure of z

avanturin [10]

9514 1404 393

Answer:

132°

Step-by-step explanation:

Name the vertex of each angle the same as the angle letter. Name the intersection of the "horizontal" and "vertical lines" point Q.

Angle ZXQ is vertical to ∠x, so is the same measure.

Angle YQX is the value that makes the sum of angles in triangle XYQ be 180°. That is ...

∠YQX = 180° -51° -57° = 72°

This is also the measure of its vertical angle in the other triangle. Angle z is the sum of that vertical angle and 60°, so we have ...

∠z = 72° +60°

∠z = 132°

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Additional comment

The relations we used are ...

vertical angles are congruent
sum of angles in a triangle is 180°
an exterior angle is equal to the sum of the remote interior angles

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