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3 years ago

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 (x-6)(x-7) a) x^2-13x-15 b) x^2-13x+42 c) x^2+x+42 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

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

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

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

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

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

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

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

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

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

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

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

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

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

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

Then the general solution to the DE is

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

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:

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

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3 years ago