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vivado [14]
3 years ago
7

Given that ∠ABC ≅ ∠DBE, which statement must be true?

Mathematics
2 answers:
Mnenie [13.5K]3 years ago
5 0

Answer:∠ABD ≅ ∠CBE   is correct statement

Step-by-step explanation:

Strike441 [17]3 years ago
3 0

Answer

∠ABD ≅ ∠CBE   is correct statement

It is given that ∠ABC ≅ ∠DBE,

Here vertex B is common for two triangles.It implies that,

∠ABCand ∠DBE are pair of vertically opposite angles. (vertically opposite angles are equal.

∠ABC ≅ ∠ABD False. (Ab is common, These angles are adjacent s angles)

∠ABD ≅ ∠CBE  True.(B is the common vertex, another pair of vertically opposite angle))

∠CBD ≅ ∠DBE Fasle  

∠CBD ≅ ∠ABC False

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3⋅f(−4)−3⋅g(−2)= -27
Reil [10]

Answer:

3⋅f(−4)−3⋅g(−2)= -27 is g = -9/2 +2f

Graph the line using the slope and y-intercept, or two points.

Slope:

6 0
3 years ago
Please help me with this
Zarrin [17]
The two end points of our line are (0,2.5) and (5,5.5). to find our slope we need  to find our n difference (higher value - lower value) and our C difference (same as n) which would be 5 and 3. now we put our cost difference over our miles difference and get our slope (5/3). this means we still need out starting point. which is a cost of 2.5. the equation is C = (slope)5/3n + (starting value)2.5
7 0
3 years ago
Read 2 more answers
The perimeter of a triangle is 86 inches. The largest side is four inches less than twice the smallest side. The third side is 1
Xelga [282]
Hello,
Assume x the smallest side.
y the third side
z the largest side
z=2x-4
y=10+x
x+y+z=86
==>x+(10+x)+(2x-4)=86
==>4x=86-10+4
==>x=20, y=30 and z=36

7 0
3 years ago
Classify ABC by its sides. Then determine whether it is a right triangle.
m_a_m_a [10]

Answer:

∴Given Δ ABC is not a right-angle triangle

a= AB = √45 = 3√5

b = BC = 12

c = AC = √45 = 3√5

Step-by-step explanation:

Given vertices are A(3,3) and B(6,9)

            AB = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2}  }

            AB = \sqrt{(9-3)^{2}+(6-3)^{2}  } = \sqrt{6^{2}+3^{2}  } =\sqrt{45}

Given vertices are  B(6,9) and C( 6,-3)

       B C = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2}  }

             =  \sqrt{(-3-9)^{2}+(6-6)^{2}  } =\sqrt{12^{2} } = 12

    BC = 12

Given vertices are  A(3,3) and C( 6,-3)

 AC = \sqrt{(6-3)^{2}+(-3-3)^{2}  } = \sqrt{9+36} = \sqrt{45}

AC² = AB²+BC²

45  = 45+144

 45  ≠ 189

∴Given Δ ABC is not a right angle triangle

 

5 0
3 years ago
Subject Algebra 1
Mashutka [201]

Answer:

10x^{2}-34

Step-by-step explanation:

Because we are told equivalent expressions for y and z we can plug those in to 2(y+z).

2((3x^{2}+x^{2}-5)+(x^{2}-12))

Then simplify by combining like terms of the expressions. Values ending in x^2 can be combined with each other.

2(5x^{2}-17)

Now we can distribute the 2 by multiplying each value in the parentheses by 2.

10x^{2}-34

6 0
3 years ago
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