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3 years ago

Combine the like terms to create an equivalent expression. 10k+1+8k+2

Mathematics

2 answers:

Margarita [4]3 years ago

8 0

10k+8k+1+2
18k+3
The answer is 18k+3

o-na [289]3 years ago

7 0

Once you combine like terms you are left with 18k +3.

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Step-by-step explanation:

yes.

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3 years ago

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3 years ago

Given: m∠A = m∠C = 90° AB ║ DC Prove: ∆ABD ≅ ∆CBD

Genrish500 [490]

Answer:

Angle-Angle-Side (AAS)

Step-by-step explanation:

Given triangles; ΔABD and ΔCBD

<A = <C = $90^{0}$ (right angle property)

AB ║ DC (given)

Then,

DA ║ CB

<B ≅ <D (congruent property)

DB ≅ BD (similarity property)

<D + <B = <B + <D (complementary angles, and property of triangles)

Therefore by Angle-Angle-Side (AAS),

∆ABD ≅ ∆CBD

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3 years ago

The triangles below are similar. Triangle A B C. Side A C is 10 and side A B is 5. Angle C is 30 degrees. Triangle D E F. Side E

Gnoma [55]

Answer:

$\triangle ABC {\displaystyle \sim } \triangle DE\ F$

$\triangle CBA {\displaystyle \sim } \triangle FED$

$\triangle BAC {\displaystyle \sim } \triangle EDF$

Step-by-step explanation:

Given

$\triangle ABC {\displaystyle \sim } \triangle DE\ F$

$AC = 10$

$AB = 5$

$\angle C = 30^\circ$

$ED = 7.5$

$DF = 25$

$\angle F =30$

$\angle E =90$

Required

Which of the options is/are true:

$\triangle CBA {\displaystyle \sim } \triangle FED$ $\triangle CBA {\displaystyle \sim } \triangle FDE$ $\triangle BAC {\displaystyle \sim } \triangle EFD$

$\triangle BAC {\displaystyle \sim } \triangle EDF$ $\triangle ABC {\displaystyle \sim } \triangle DE\ F$ $\triangle ABC {\displaystyle \sim } \triangle DFE$