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3 years ago

find the value of the polygon. the polygon has 6 interior angles 2 of them are y and 4 of them are (2Y-20)

1 answer:

3 0

First we need to determine what the 6 angles must add to. Turns out we use this formula

S = 180(n-2)

where S is the sum of the angles (result of adding them all up) and n is the number of sides. In this case, n = 6. So let's plug that in to get

S = 180(n-2)
S = 180(6-2)
S = 180(4)
S = 720

The six angles, whatever they are individually, add to 720 degrees. The six angles are y, y, 2y-20, 2y-20, 2y-20, 2y-20, 

They add up and must be equal to 720, so let's set up the equation to get...

(y)+(y)+(2y-20)+(2y-20)+(2y-20)+(2y-20) = 720

Let's solve for y

y+y+2y-20+2y-20+2y-20+2y-20 = 720

10y-80 = 720

10y-80+80 = 720+80

10y = 800

10y/10 = 800/10

y = 80

Now that we know the value of y, we can figure out the six angles

angle1 = y = 80 degrees
angle2 = y = 80 degrees
angle3 = 2y-20 = 2*80-20 = 140 degrees
angle4 = 2y-20 = 2*80-20 = 140 degrees
angle5 = 2y-20 = 2*80-20 = 140 degrees
angle6 = 2y-20 = 2*80-20 = 140 degrees

and that's all there is to it

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Center: (9,0), Radius: 5√3

densk [106]

Answer:

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Step-by-step explanation:

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3 years ago

Sekhar gives a quarter of his sweets to Renu and then gives 5 sweets to Raji. He has

Feliz [49]

Answer:

16 sweets

Step-by-step explanation:

Number of sweet that Sekhar had = x

Sweets given to Renu = (1/4) of x = $\frac{1}{4}x$

Sweets given to Raji = 5

Remaining sweets = 7

$x = \frac{1}{4}x +5 + 7\\\\x -\frac{1}{4}x = 12\\\\\frac{4*x}{1*4}x-\frac{1}{4}x = 12\\\\\frac{4x}{4}-\frac{1}{4}x= 12\\\\\frac{3x}{4}=12\\\\x = 12*\frac{4}{3}\\\\x=4*4\\\\x=16$

6 0

2 years ago

Find the absolute maximum and absolute minimum values of the function f(x, y) = x 2 + y 2 − x 2 y + 7 on the set d = {(x, y) : |

dsp73

Looks like $f(x,y)=x^2+y^2-x^2y+7$ .

$f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1$

$f_y=2y-x^2=0\implies2y=x^2$

If $x=0$ , then $y=0$ - critical point at (0, 0).
If $y=1$ , then $x=\pm\sqrt2$ - two critical points at $(-\sqrt2,1)$ and $(\sqrt2,1)$

The latter two critical points occur outside of $D$ since $|\pm\sqrt2|>1$ so we ignore those points.

The Hessian matrix for this function is

$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}$

The value of its determinant at (0, 0) is $\det H(0,0)=4>0$ , which means a minimum occurs at the point, and we have $f(0,0)=7$ .

Now consider each boundary:

If $x=1$ , then

$f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4$

which has 3 extreme values over the interval $-1\le y\le1$ of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

If $x=-1$ , then

$f(-1,y)=8-y+y^2$

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

If $y=1$ , then

$f(x,1)=8$

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

If $y=-1$ , then

$f(x,-1)=2x^2+8$

which has 3 extreme values, but the previous cases already include them.

Hence $f(x,y)$ has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

3 0

2 years ago

Solving Quadratic Equations with Complex SolutionsThe discriminant: D=b^2-4acQuestion:5x^2-2x+1=0

fomenos

• The value of the discriminant ,D= -16

• The solution to the quadratic equation is

$x=\frac{1+2i}{5}\text{ or }\frac{1-2i}{5}$

Step - by - Step Explanation

What to find?

• The discriminant d= b² - 4ac

• The solution to the quadratic equation.

Given:

5x² - 2x + 1=0

Comparing the given equation with the general form of the quadratic equation ax² + bx + c=0

a=5 b=-2 and c=1

Uisng the quadratic formula to solve;

$x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

The discriminant D=b² - 4ac

Substitute the values into the discriminant formula and simplify.

D = (-2)² - 4(5)(1)

D = 4 - 20

D = -16

We can now proceed to find the solution of the quadratic equation by substituting into the quadratic formula;

$x=\frac{-(-2)\pm\sqrt[]{-16}}{2(5)}$

Note that:

√-1 = i

$x=\frac{2\pm\sqrt[]{16\times-1}}{10}$

$x=\frac{2\pm\sqrt[]{16}\times\sqrt[]{-1}}{10}$ $x=\frac{2\pm4i}{10}$ $x=\frac{2}{10}\pm\frac{4i}{10}$ $x=\frac{1}{5}\pm\frac{2}{5}i$ $x=\frac{1\pm2i}{5}$

That is;

$\text{Either x=}\frac{1+2i}{5}\text{ or x=}\frac{1-2i}{5}$

7 0

10 months ago

Latitude and longitude describe locations on the Earth with respect to the equator and prime meridian. The table shows the

Contact [7]

Answer:

Step-by-step explanation:

Given that latitude and longitude describe locations on the Earth with respect to the equator and prime meridian.

Latitude High Temp

("N)

12 53

16 41

30 67

36 63

32 70

11 58

10 61

33 67

30 72

Slope 0.732

Int -21.56266667

Regression line is y = 0.732x-21.563 where x represents the latitude and y high temperature

Since slope = 0.732, we get temp increases by 0.8 degree for 1 degree increasenorth in latitude

Hence option c is right

7 0

2 years ago

Read 2 more answers