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lapo4ka [179]
3 years ago
6

Find the length AB i'm your answer to the tenth

Mathematics
1 answer:
natali 33 [55]3 years ago
7 0

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️AB =  \sqrt{ ({8 - 4})^{2} +  ({9 - 1})^{2}  }

AB =  \sqrt{ {4}^{2} +  {8}^{2}  }

AB =  \sqrt{16 + 64}

AB =  \sqrt{80}

AB = 8.94

AB = 8.9♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

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Complete the ordered pair for the given equation. (_,-1);y=-3/5x-7
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So for this problem, we're given the y-coordinate and we can plug that in for y in the equation. So let's do that

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Question 18(Multiple Choice Worth 1 points) (05.02 MC) The figure shows a parallelogram inside a rectangle outline: 1 foot B ---
lesya692 [45]

Answer:

1/9

Step-by-step explanation:

You find the area of the rectangle first (1/2 by 1/3)

then find the area excluding the parallelogram (the two triangles outside, which would be 1/6 by 1/3 since both are of the same side measurement)

(sorry if this is hard for you to understand by the way)

Then subtract the answer from step 2 from step 1 and voila, you get your answer

7 0
3 years ago
Can someone please help me
Afina-wow [57]

Answer:

Step-by-step explanation:

1) Statements                         Reasons

ABCD is a trapezoid               Given

AD ║ BC                                  bases of trapezoid are parallel

∠AED = ∠CEB                          vertically opposite angles are equal

∠EBC = ∠ADE          alternate interior angles are equal and BD is transversal                

∠ECB = ∠EAD          alternate interior angles are equal and AC is transversal

ΔAED ~ ΔCEB            ∠ECB = ∠EAD & ∠AED = ∠BEC & ∠EBC = ∠ADE

                                all angles are same and their shape is same, so similar

2) hope you can write 2nd one as above two column proof

T is the mid point of QR

U is the mid point of QS

V is the mid point of RS

Definition: The segment of line joining the middle points of two sides of a triangle is called middle segment.

Inference: A middle segment of a triangle is parallel to the third side and its lengths is half the third side's.

so here TU ║ RV and TU = RV  

UV ║ TR and UV = TR

so TUVR becomes a parallelogram

∠TUV = ∠TRV -------->  condition 1

same goes with parallelogram TQUV

∠TQU = ∠UVT ----------> condition 2

same goes with parallelogram TUSV

∠UTV = ∠USV ------------> condition 3

from the above 3 conditions we can say

ΔQRS ~ ΔVUT

3) see the below figure for graph

in both triangles ∠B is common

AC and TS are parallel lines

BC  and BA are transversals

∠C ∠S are equal --> corresponding angles

∠A ∠T are equal ---> corresponding angles

so ΔABC & ΔTBS are similar

3 0
3 years ago
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