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Ray Of Light [21]
3 years ago
5

What would the sum of the interior angles of a 14 sided shape be?

Mathematics
2 answers:
Yanka [14]3 years ago
7 0

Answer:  D)  2160

Work Shown:

S = sum of the interior angles of a polygon with n sides

S = 180(n-2)

S = 180(14-2)

S = 180*12

S = 2160

labwork [276]3 years ago
3 0

Answer:

the sum of interior angles of a 14 sided shape be 2160

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Which of the following trigonometric ratios are correct?
den301095 [7]

Hi there! In this problem, you should have the knowledge of three basic Trigonometric Ratio.

  • sinA = opposite/hypotenuse
  • cosA = adjacent/hypotenuse
  • tanA = opposite/adjacent

Now that we know three basic ratio. Let's check each choices!

  • sin50° = d/c

This choice is wrong because we focus on the 50° angle. When we focus on 50°, sin50° should be d/x and not d/c.

  • sin50° = c/x

This choice is also wrong because in ratio, it's cos50° that adjacent/hypotenuse.

  • tan50° = d/c

This choice is correct! As ratio states, tanA = opposite/adjacent.

  • tan50° = x/c

This choice is wrong. x/c is a reciprocal of cosine which is 1/cos. We call the reciprocal of cosine as secant or sec in short.

  • cos50° = x/d

This choice is wrong as x/d is a reciprocal of sine which is 1/sin. We call the reciprocal of sine as cosecant or cosec/csc in short.

  • cos50° = c/x

This choice is right by the ratio. Nothing really much to explain since we follow by ratio that is defined.

Answer

  • tan50° = d/c
  • cos50° = c/x

Questions can be asked through comment.

Furthermore, tan also has its reciprocal form itself which is called cotangent also known as cot in short.

Hope this helps, and Happy Learning! :)

8 0
3 years ago
Distribute <br>(x-6)(x-7)<br><br>a) x^2-13x-15<br>b) x^2-13x+42<br>c) x^2+x+42<br>d) x^2+x-15​
Sidana [21]

Answer:

B. x² - 13x + 42

Step-by-step explanation:

7 0
3 years ago
which correctly shows how to use the GCF and the distributive property to find an expression equivalent to 24+44
Sergeu [11.5K]
I would say it is 4(6+11) that or 72 is the answer
7 0
3 years ago
Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x​
Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

<em>r</em> ² - 3<em>r</em> + 2 = (<em>r</em> - 2) (<em>r</em> - 1) = 0

with roots <em>r</em> = 2 and <em>r</em> = 1, so the characteristic solution is

<em>y</em> (char.) = <em>C₁</em> exp(2<em>x</em>) + <em>C₂</em> exp(<em>x</em>)

For the <em>ansatz</em> particular solution, we might first try

<em>y</em> (part.) = (<em>ax</em> + <em>b</em>) + (<em>cx</em> + <em>d</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)

where <em>ax</em> + <em>b</em> corresponds to the 2<em>x</em> term on the right side, (<em>cx</em> + <em>d</em>) exp(<em>x</em>) corresponds to (1 + 2<em>x</em>) exp(<em>x</em>), and <em>e</em> exp(3<em>x</em>) corresponds to 4 exp(3<em>x</em>).

However, exp(<em>x</em>) is already accounted for in the characteristic solution, we multiply the second group by <em>x</em> :

<em>y</em> (part.) = (<em>ax</em> + <em>b</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)

Now take the derivatives of <em>y</em> (part.), substitute them into the DE, and solve for the coefficients.

<em>y'</em> (part.) = <em>a</em> + (2<em>cx</em> + <em>d</em>) exp(<em>x</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)

… = <em>a</em> + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)

<em>y''</em> (part.) = (2<em>cx</em> + 2<em>c</em> + <em>d</em>) exp(<em>x</em>) + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

… = (<em>cx</em> ² + (4<em>c</em> + <em>d</em>)<em>x</em> + 2<em>c</em> + 2<em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

Substituting every relevant expression and simplifying reduces the equation to

(<em>cx</em> ² + (4<em>c</em> + <em>d</em>)<em>x</em> + 2<em>c</em> + 2<em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

… - 3 [<em>a</em> + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)]

… +2 [(<em>ax</em> + <em>b</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)]

= 2<em>x</em> + (1 + 2<em>x</em>) exp(<em>x</em>) + 4 exp(3<em>x</em>)

… … …

2<em>ax</em> - 3<em>a</em> + 2<em>b</em> + (-2<em>cx</em> + 2<em>c</em> - <em>d</em>) exp(<em>x</em>) + 2<em>e</em> exp(3<em>x</em>)

= 2<em>x</em> + (1 + 2<em>x</em>) exp(<em>x</em>) + 4 exp(3<em>x</em>)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

<em>x</em> : 2<em>a</em> = 2

1 : -3<em>a</em> + 2<em>b</em> = 0

exp(<em>x</em>) : 2<em>c</em> - <em>d</em> = 1

<em>x</em> exp(<em>x</em>) : -2<em>c</em> = 2

exp(3<em>x</em>) : 2<em>e</em> = 4

Solving the system gives

<em>a</em> = 1, <em>b</em> = 3/2, <em>c</em> = -1, <em>d</em> = -3, <em>e</em> = 2

Then the general solution to the DE is

<em>y(x)</em> = <em>C₁</em> exp(2<em>x</em>) + <em>C₂</em> exp(<em>x</em>) + <em>x</em> + 3/2 - (<em>x</em> ² + 3<em>x</em>) exp(<em>x</em>) + 2 exp(3<em>x</em>)

4 0
3 years ago
What’s the answer to this?
Elina [12.6K]

Step-by-step explanation:

Line is passing through the points:

(-2,\:3)=(x_1,\:y_1) \:\&\: (2,\:5)=(x_2 ,\:y_2)

Equation of line in two point form is given as:

\frac{y -y_1 }{y_1 -y_2 } = \frac{x -x_1 }{x_1 -x_2 } \\ \\ \therefore \frac{y -3}{ 3 -5} = \frac{x -(-2) }{-2 -2} \\ \\ \therefore \frac{y -3}{ - 2} = \frac{x -(-2) }{-4} \\ \\ \therefore \frac{y - 3}{1} = \frac{x +2}{2} \\ \\ \therefore \: y-3= \frac{x}{2} + \frac{2}{2} \\ \\\therefore \: y= \frac{x}{2} + 1 +3\\ \\ \huge \red{ \boxed{\therefore \: y= \frac{1}{2} \:x+ 4}}\\ is \: the \: required \: equation \: of \: line.

4 0
3 years ago
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