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3 years ago

Find the sum of the interior angles of a quadrilateral with 15 sides?

Mathematics

2 answers:

OlgaM077 [116]3 years ago

6 0

Answer:

2,340

Step-by-step explanation:

Using Formula

=> $(n-2)*180$

Where n is the number of sides. Here, no. of sides is 15

=> $(15-2)*180$

=> $13*180$

=> 2,340

Crank3 years ago

5 0

Answer:

2340°

Step-by-step explanation:

To find the sum of interior angles of any shape, use the formula: 180(n - 2), where n is how many sides there are. So,

180(15 - 2)

180(13)

2340

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Andreyy89

Let ${s_n}_{n\in\Bbb N}$ be a sequence that converges to $L$ . This means for any $\varepsilon>0$ , there is some $N$ such that $|s_n-L|$ for all $n>N$ . From this inequality we see that $|(s_n-L)-0|$ , so it follows that $s_n-L\to0$ .

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2 years ago

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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3 years ago

Slove algebraically 3x - 4y = -24 x +4y=8

Vanyuwa [196]

Solve algebrically 3x - 4y = -24 and x + 4y = 8 is x = -4 and y = 3

Solution:

We have been given two equations which are as follows:

3x - 4y = -24 ----- eqn 1

x + 4y = 8 -------- eqn 2

We have been asked to solve the equations which means we have to find the value of ‘x’ and ‘y’.

We rearrange eqn 2 as follows:

x + 4y = 8

x = 8 - 4y ------eqn 3

Now we substitute eqn 3 in eqn 1 as follows:

3(8 - 4y) -4y = -24

24 - 12y - 4y = -24

-16y = -48

y = 3

Substitute "y" value in eqn 3. Therefore the value of ‘x’ becomes:

x = 8 - 4(3)

x = 8 - 12 = -4

Hence on solving both the given equations we get the value of x and y as -4 and 3 respectively.

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2 years ago

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Vinil7 [7]

C

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3 years ago

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Find gradient and y-intercept of Y=3x-5

NemiM [27]

Answer:

Slope m (Gradient) = 3

y-intercept: (0, -5)

General Formulas and Concepts:

Algebra I

Slope-Intercept Form: y = mx + b

m - slope

b - y-intercept

Step-by-step explanation:

Step 1: Define

y = 3x - 5

Step 2: Break function

Slope m = 3

y-intercept b = -5

8 0

3 years ago

Read 2 more answers