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Gala2k [10]
2 years ago
14

Can anyone help me with this question please

Mathematics
1 answer:
klemol [59]2 years ago
8 0

Answer:

  7.  ∠CBD = 100°

  8.  ∠CBD = ∠BCE = 100°; ∠CED = ∠BDE = 80°

Step-by-step explanation:

7. We presume the angles at A are congruent, so that each is 180°/9 = 20°.

Then the congruent base angles of isosceles triangle ABC will be ...

  ∠B = ∠C = (180° -20°)/2 = 80°

The angle of interest, ∠CBD is the supplement of ∠ABC, so is ...

  ∠CBD = 180° -80°

  ∠CBD = 100°

__

8. In the isosceles trapezoid, base angles are congruent, and angles on the same end are supplementary:

  ∠CBD = ∠BCE = 100°

  ∠CED = ∠BDE = 80°

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Answer:

okay look at it this way:

Step-by-step explanation:

when you use a mirror and you look at the shape on both sides then you will see the -x which is basically on the same side but just a minus instead. Soo let's say you have to graph a coordinate of (2,4) y=2 and x=4 si you have to graph the normal plots and when you put it in the opposite sides the thing you only did was you just made a reflection of the shape just negative instead. hope you understand what I'm saying.

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What is the value of p if the line that passes through (2,-4) and (p,8) has a slope of 1/2? ​
Irina18 [472]

Answer:

Step-by-step explanation:

(2,-4)....x1= 2 and y1 = -4

(p,8).....x2 = p and y2 = 8

slope(m) = 1/2

now use the slope formula (y2 - y1) / (x2 - x1) and sub in what we know...

slope(m) = (y2 - y1) / (x2- x1)

1 / 2 = (8 - (-4) / (p - 2)

1 / 2 = (8 + 4) / (p - 2)

1/2 = 12 / (p - 2) .....now multiply both sides by (p - 2)

1/2(p - 2) = 12

1/2p - 1 = 12

1/2p = 12 + 1

1/2p = 13

p = 13 / (1/2)

p = 13 * 2/1

p = 26

so the value of p is 26

6 0
3 years ago
Find the other two vertices of a square with one vertex (0, 0) and another vertex (4, 2). Can you find another answer?
Margarita [4]

It is given that two vertices of square are (0,0) and (4,2).

Now the problem is that you haven't given that whether these two vertices are adjacent vertices or opposite vertices of the square.

1. By Supposing that these two are adjacent vertices of Square

The third vertex will be at (-4,2) which lies in third quadrant.

Suppose the coordinate of fourth vertex be (x,y).

Mid point of line joining (4,2) and (-4,2) is{ [4+(-4)]/2,(2+2)/2} is (0,2).

Mid point of line joining (x,y) and (0,0) is (x/2,y/2).

Since diagonals of square bisect each other,

∵ x/2=0

⇒x=0

and

y/2=2

⇒y=4

So, The Coordinate of  fourth vertex is (0,4).

Now coming back to second condition if these are two opposite vertex of Square.

Let the third coordinate be (a,b).

Length of diagonal=\sqrt{(4-0)^2+(2-0)^2}=\sqrt20=2\sqrt5

Now,let side of Square be A.

Then length of Diagonal of square =√2 A

⇒√2 A=2√5

⇒A =√10

As third vertex is (a,b).

Using distance formula

a² + b²=10  -------------(1)

(a-4)²+ (b-2)²=10  --------------(2)

Solving expression (1) and (2), we get

⇒a²+ b²=(a-4)² +(b-2)²

⇒2a + b =5

⇒b=5-2a

Putting the value of b in (1),we get

⇒a² +(5-2a)²=10

⇒a²+25+4a²-20a =10

⇒5a²-20a+15=0

⇒a² - 4a + 3=0

Splitting the middle term,we get

⇒(a-3)(a-1)=0

⇒a=3  ∧  a=1

we get b=5-2×1=3 and b=5-2×3=5-6=-1

So,the other vertex are (1,3) and(3,-1).





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2 years ago
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