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2 years ago

2h + (6 + h) + 6 ∙ 2

Mathematics

2 answers:

ANEK [815]2 years ago

6 0

Answer: 3h+18

Step-by-step explanation:

2h+6+h+12

add like terms (add all the the h’s) which equals 3h and add 6+12 which equals 18 so your answer is 3h+18

Readme [11.4K]2 years ago

4 0

3h+18 is the correct answer:)

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natita [175]

Answer:

The correct answer is: $\Delta BDE \cong \Delta BFK$ by rule ASA rule of congruence.

Step-by-step explanation:

First let us prove $\Delta BDE \cong \Delta BFK$ by rule ASA (rule of congruence).

Congruent side:

$\overline{BD} \cong \overline{BF}$ (Given)

Congruent angles:

1. By definition of perpendicular,

$\angle{BFK} = 90 \textdegree \ (Since \ \overline{FK} \ is \ perpendicular \ to \ \overline{AB} \ (\overline{FK} \perp \overline{AB} =Given))$

Also,

$\angle{BDE} = 90 \textdegree \ (Given)$

Therefore,

$\angle{BDE} \cong \angle{BFK}$

or you can say,

$\angle{D} \cong \angle{F}$

2. Common angle between $\Delta BDE \ and \ \Delta BFK$ is $\angle{B}$

In a nutshell, in $\Delta BDE$ ,

$\angle{B}$ (Angle)

$\overline{BD}$ (Side)

$\angle{BDE}$ (Angle)

are congruent to the following angle, side and angle of $\Delta BFK$ :

$\angle{B}$ (Angle)

$\overline{BF}$ (Side)

$\angle{BFK}$ (Angle)

Therefore, by ASA rule of congruence, we can say $\Delta BDE \cong \Delta BFK$ .

Since both triangles are congruent, the sides $\overline{ED}$ and $\overline{FK}$ are also congruent.

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2 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

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Step-by-step explanation:

4.375 times 10 to the -8

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2 years ago

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Answer:

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Step-by-step explanation:

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