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BaLLatris [955]
3 years ago
13

Solve -6x+19 = x-2. Check for extraneous solutions.

Mathematics
2 answers:
dmitriy555 [2]3 years ago
6 0

Answer:

x=3

Step-by-step explanation:

Move the variables to the left-hand side and change its sign. Move constant to the right-hand side and change its sign.

-6x-x=-2-19

Collect like terms

-7x=-21

Divide both sides of the equation by -7

x=3

Hope this helpsʕ•ᴥ•ʔ

Eddi Din [679]3 years ago
3 0

Answer:

C. x = 3

Step-by-step explanation:

Step 1:

- 6x + 19 = x - 2

Step 2:

- 7x + 19 = - 2

Step 3:

- 7x = - 21

Step 4:

21 = 7x

Answer:

3 = x

Hope This Helps :)

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Can someone please help me
Afina-wow [57]

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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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