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3 years ago

7

What is 2462 divided by 2

2 answers:

Darina [25.2K]3 years ago

8 0

2462 divided by 2 is 1231. <u />A easier way to do this simple problem without doing the math is 2 divided by 2 equals 1. 4 divided by 2 equals 2. 6 divided by 2 equals 3 and 2 divided by 2 equals 1. You put it together and you get 1231. I'm sorry because i cant show the work<u>
</u>

stiv31 [10]3 years ago

3 0

Dividing by 2 is the same as multiplying by 1/2.

2462 * 1/2 = 1231

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Given that P = (-4, 11) and Q = (-5, 8), find the component form and magnitude of vector QP

Svetach [21]

The given points are
P = (-4,11)
Q = (-5,8)

The x-component of vector QP is
-4 - (-5) = 1
The y-component of vector QP is
11 - 8 = 3

The vector QP is
(1,3) or
$\vec{QP} = \hat{i} + 3\hat{j}$

The magnitude of the vector is
√(1² + 3²) = √(10)

Answer:
$\vec{QP} = \hat{i}+3\hat{j} \,\, or \,\, (1,3)$
The magnitude is √(10).

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3 years ago

What is 9k + 1 = ─9 + 7k solved for the variable???

m_a_m_a [10]

Answer:

k = -5

Step-by-step explanation:

The variable is k

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-7k -1

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4 0

3 years ago

Help please and don't delete the question

Ahat [919]

Answer:

x = 2

Step-by-step explanation:

8(4-x) = 7x + 2

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2 years ago

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Question : In the given figure , ∆ APB and ∆ AQC are equilateral triangles. Prove that PC = BQ.

lorasvet [3.4K]

Answer:

See Below.

Step-by-step explanation:

We are given that ΔAPB and ΔAQC are equilateral triangles.

And we want to prove that PC = BQ.

Since ΔAPB and ΔAQC are equilateral triangles, this means that:

$PA\cong AB\cong BP\text{ and } QA\cong AC\cong CQ$

Likewise:

$\angle P\cong \angle PAB\cong \angle ABP\cong Q\cong \angle QAC\cong\angle ACQ$

Since they all measure 60°.

Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:

$m\angle PAC=m\angle PAB+m\angle BAC$

Likewise:

$m\angle QAB=m\angle QAC+m\angle BAC$

Since ∠QAC ≅ ∠PAB:

$m\angle PAC=m\angle QAC+m\angle BAC$

And by substitution:

$m\angle PAC=m\angle QAB$

Thus:

$\angle PAC\cong \angle QAB$

Then by SAS Congruence:

$\Delta PAC\cong \Delta BAQ$

And by CPCTC:

$PC\cong BQ$

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3 years ago

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egoroff_w [7]

Answer:

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Step-by-step explanation:

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