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Blababa [14]
3 years ago
14

Describe all situation where vertical angles are also supplementary

Mathematics
2 answers:
timofeeve [1]3 years ago
6 0

Answer:

if  vertical angles are right angles  then the angles are supplementary.

Step-by-step explanation:

Vertical angles: When two lines are intersect to each other then the opposite angles made at the point of intersection are called vertical angles.

Vertical angles are always equal in measurement.

Supplementary angles: The angles whose sum is 180 degrees are called supplementary angle.

Let x and y are vertical angles

Then, x=y

If angles x and y are supplementary

x+y=180^{\circ}

x+x=180^{\circ}

2x=180^{\circ}

x=\frac{180}{2}=90^{\circ}

x=y=90^{\circ}

When x and y=90 degrees then the angle x and y are supplementary.

Therefore, we can say that if vertical angles are right angles  then the angles are supplementary.

Nataly_w [17]3 years ago
3 0
Vertical angles are always the same size. if they are supplementary the measures must add to 180. 180/2=90. vertical angles are supplementary only when they are right angles, our 90°
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Zolol [24]

Answer:

x=0,x=-2

Step-by-step explanation:

From the question we are told that:

y=2x^5+5x^4-19

Generally the equation if differentiated is mathematically given by

y'=10x^4+20x^3-0

Where

y'=0

10x^4+20x^3=0

Factorizing,We have

x=0,x=-2

Therefore

The critical points are

x=0,x=-2

4 0
2 years ago
Find the absolute maximum and minimum values of f(x,y)=xy−4x in the region bounded by the x-axis and the parabola y=16−x2.
skad [1K]

Answer:

The absolute maximum and minimum is 20\; \text{and} -20.

Step-by-step explanation:

We first check the critical points on the interior of the domain using the

first derivative test.

f_x=y-4=0

f_y=x=0

The only solution to this system of equations is the point (0, 4), which lies in the domain.

f_{xx}=0, \;f_{yy}=0\; \text{and}\; f_{xy}=-1

\Rightarrow f_{xx}f_{yy}-f_{xy}=o-1=-1

\therefore (0,4) is a saddle point.

Boundary points -  (4,0),  (-4,0), (0,16)

Along boundary  y=16-x^2

   f=x(16-x^2)-4x

=16x-x^3-4x

\Rightarrow f^'=16-3x^2-4=0

\Rightarrow 3x^2=12

\Rightarrow x=\pm2,\;\;y=14

Values of f(x) at these points.

\begin{array}{}(4,0)=-16\\(-4,0)=16\\(0,16)=0\\(2,14)=20\\(-2,14)=-20\end{array}

Therefore, the absolute maximum and minimum is 20\; \text{and} -20.

6 0
3 years ago
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svetlana [45]

Answer:

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8 0
3 years ago
4x – 3y=8<br> 2. 5x – 2y = -11
AlladinOne [14]

Answer:

x = -7

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Step-by-step explanation:

The given system is:

4x-3y=8

5x-2y=-11

We make x the subject in the top equation to get:

x =  \frac{3}{4}y + 2

Plug this expression for x in the bottom equation to get

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We expand to obtain

\frac{15}{4} y + 10 - 2y =  - 11

Multiply through by 4

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Group similar terms

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7y =  - 84

y =  -  \frac{84}{7}  =-12

This means

x =  -12\times  \frac{3}{4}  + 2

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6 0
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8+-10=-2  you multiply and combine the variables and numbers.
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