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mrs_skeptik [129]
3 years ago
15

Need help any body please help

Mathematics
1 answer:
Tcecarenko [31]3 years ago
4 0
Attached a picture
hope it helps :)

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Write an equation for the perpendicular bisector of the line segment joining the two points.
Leona [35]
Points (1, 7)  and  (-3, 2)

Slope for a line between (x₁, y₁) and (x₂, y₂) , m = (y₂ -y₁) / (x₂- x₁)

The slope for the line joining the two points =  (2 - 7) / (-3 - 1) = -5/-4

Slope = 5/4

Hence the perpendicular bisector would have a slope of -1/(5/4) = -4/5

By condition of perpendicularity

For points (1, 7)  and  (-3, 2),

Formula for midpoints for (x₁, y₁) and (x₂, y₂) is ((x₁ +x₂)/2 , (y₁+ y₂)/2)

Midpoint for (1, 7)  and  (-3, 2) = ((1+ -3)/2 , (7+2)/2) = (-2/2, 9/2)

= (-1, 9/2)

Since the slope of perpendicular bisector is -4/5 and passes through the midpoint (-1, 9/2)

Equation  y - y₁ = m (x - x₁)

                y - 9/2 = (-4/5) (x - -1)
           
                y - 9/2 = (-4/5)(x + 1)

               5(y - 9/2) = -4(x + 1)

               5y - 45/2 = -4x - 4

                 5y =  -4x - 4 + 45/2
           
                5y + 4x = 45/2 - 4

                  5y + 4x = 22  1/2  - 4 =  18 1/2

                  5y + 4x =  37/2

                  10y + 8x = 37

The equation of the line to perpendicular bisector is 10y + 8x = 37
7 0
4 years ago
Verify the identity
Igoryamba

Answer:

We have to prove

sin⁡(α+β)-sin⁡(α-β)=2 cos⁡ α sin ⁡β

We will take the left hand side to prove it equal to right hand side

So,

=sin⁡(α+β)-sin⁡(α-β)      Eqn 1

We will use the following identities:

sin⁡(α+β)=sin⁡ α cos⁡ β+cos⁡ α sin⁡ β

and

sin⁡(α-β)=sin⁡ α cos ⁡β-cos ⁡α sin ⁡β

Putting the identities in eqn 1

=sin⁡(α+β)-sin⁡(α-β)

=[ sin⁡ α cos ⁡β+cos⁡ α sin⁡ β ]-[sin⁡ α cos ⁡β-cos ⁡α sin ⁡β ]

=sin⁡ α cos⁡ β+cos⁡ α sin ⁡β- sin⁡α cos⁡ β+cos ⁡α sin ⁡β

sin⁡α cos⁡β will be cancelled.

=cos⁡ α sin ⁡β+ cos ⁡α sin ⁡β

=2 cos⁡ α sin ⁡β  

Hence,

sin⁡(α+β)-sin⁡(α-β)=2 cos ⁡α sin ⁡β

8 0
3 years ago
Prove that DE is parallel to BC. <br> Please help, will award brainliest.
Gennadij [26K]

Answer:

see explanation

Step-by-step explanation:

Parallel lines have equal slopes.

To find D and E use the midpoint formula

Given endpoints (x₁, y₁ ) and (x₂, y₂ ) then the midpoint is

( \frac{x_{1}+x_{2}  }{2}, \frac{y_{1}+y_{2}  }{2} )

Here (x₁, y₁ ) = A(4, 6) and (x₂, y₂ ) = B(2, - 2) , then

D = (\frac{4+2}{2}, \frac{6-2}{2} ) = (3, 2 ) and

let (x₁, y₁ ) = B(2, - 2\frac{-4+2}{-2-2} ) and (x₂, y₂ ) = C(- 2, - 4 ), then

E = ( \frac{4-2}{2}, \frac{6-4}{2} ) = (1, 1 )

Use the slope formula to find slopes of DE and BC

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = D(3, 2) and (x₂, y₂ ) = E(1, 1), then

m_{DE} = \frac{1-2}{1-3} = \frac{-1}{-2} = \frac{1}{2}

Repeat with (x₁, y₁ ) = B(2, - 2) and (x₂, y₂ ) = C(- 2, - 4), then

m_{BC} = \frac{-4+2}{-2-2} = \frac{-2}{-4} = \frac{1}{2}

Since the slopes are equal then DE and BC are parallel lines

6 0
3 years ago
Read 2 more answers
The pyramid shown has a square base that is 24 centimeters on each side. The slant height is 16 centimeters. What is the lateral
Aleksandr-060686 [28]
Check the picture below.

the "lateral" area, or "sides" area, is just the area of all the four triangular faces, and it doesn't include the bottom or base of the pyramid.

however, notice, each triangular face is really just a triangle with a base of 24, and a height of 16.

\bf \left[\frac{1}{2}(\stackrel{b}{24})(\stackrel{h}{16}) \right]+\left[\frac{1}{2}(\stackrel{b}{24})(\stackrel{h}{16}) \right]+\left[\frac{1}{2}(\stackrel{b}{24})(\stackrel{h}{16}) \right]+\left[\frac{1}{2}(\stackrel{b}{24})(\stackrel{h}{16}) \right]&#10;\\\\\\&#10;\textit{or just }\qquad 4\left[\frac{1}{2}(\stackrel{b}{24})(\stackrel{h}{16}) \right]\impliedby \textit{lateral area of the pyramid}

6 0
3 years ago
Read 2 more answers
Find the surface area of the come in terms of pi
ira [324]

Answer:

1157.91 or rounded 1157.9 or 1158

Step-by-step explanation:

π • r^2 • (h/3)

π • 10^2 • 11/3

8 0
3 years ago
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