Answer:
The new shape will be like the one that i have drawn for you , i think .
Step-by-step explanation:
add two ( to base* from both sides of it * )
delete one ( from hights )
Hope it was useful ;)
Help me, please ( 10 points )
2 answers:
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Answer:
9.85
Step-by-step explanation:
I hate these, LOL
We're going to draw a triangle on the graph paper to solve this. (in theory)
The top point is at 2 on the x axis, and the bottom is at 6. That's a difference of 4. Draw a line from the top dot over 4 spaces, so a line dropped from there could go straight to the bottom dot. We'll call this line/side of the triangle a, and a = 4.
Then we drop the line from there (6, 4) down to the second dot at (6, -5). That line is 9 places long. We'll call this line/side of the triangle b, and b = 9.
(This is so much easier to draw than type, I'll attach a picture).
Then we connect the two dots diagonally to form line/side c of our triangle. This is the length we don't know.
The pythagorean theorem says a^2 + b^2 = c^2
So a = 4, b = 9, c is unknown for our triangle
(4)(4) + (9)(9) = c^2
16 + 81 = 97
The square root of 97 = 9.8488, so if they have you round, I would say 9.85
Sorry if this just confused you more.
Answer:
Step-by-step explanation:
Given
See attachment for question
Required
Solve
We have:
Take LCM and solve
Answer:
∠EGB and ∠AGF are congruent;
Transitive Property of Equality
Step-by-step explanation:
The vertical angles theorem is about the angles that are opposite to each other. These angles are formed when two lines cross each other
Vertical angles are congruent i.e their measures are equal
Look at the attached graph
Given:
AB // CD
E , G , H , F are col-linear
The line which contains points E , G , H , F is a transversal for the parallel lines AB and CD
∵ AB and EF intersected at point G
∴ m∠EGB = m∠AGF ⇒ by the vertical angles theorem
When lines AB and EF intersected at G they formed opposite angles like angles EGB and AGF, then angles EGB and AGF are congruent ( vertical angle theorem)
∵ AB // CD and EF is a transversal
∴ m∠AGF = m∠CHF (corresponding angles theorem)
Transitive Property of Equality: if one angle is equal to two other angles, then the two other angles are equal
Therefore, If a = b and b = c, then a = c
∵ m∠EGB = m∠AGF
∵ m∠AGF = m∠CHF
∴ m∠EGB = m∠CHF (transitive property theorem)
∠EGB and ∠AGF are congruent; Transitive Property of Equality
