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4 years ago

The problem is:

Mathematics

2 answers:

strojnjashka [21]4 years ago

6 0

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A is 2x^2- 4xy -x +y = 0
B is (2x-1)(x+2y)

Use foil method.
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Alinara [238K]4 years ago

4 0

The answer for letter a is 2x^2- 4xy -x +y = 0
the answer for letter b is (2x-1)(x+2y) this is just the same as the answer but it is in factored form. You can try it by using foil method.

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What is the value of x that makes PQ || RS

jarptica [38.1K]

Answer:

x = 55

Step-by-step explanation:

These are alternate exterior angles and alternate exterior angles are equal when the lines are parallel

3x-65 = 2x-10

Subtract 2x from each side

3x-2x -65 = 2x-2x-10

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Add 65 to each side

x-65+65 = -10+65

x = 55

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3 years ago

Three billion sixty million sixty six thousand nine hundred

Anestetic [448]

Answer:

3,000,000,000

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Step-by-step explanation:

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3 years ago

Two buildings are facing each other on a road of width 12 metre . From the top of the first building, which is 10 metre high , t

Karo-lina-s [1.5K]

Answer:

$\displaystyle 10 + 12\sqrt{3}$ metres

Step-by-step explanation:

You are finding the height of the building with an angle of elevation, therefore we need to solve for EC to add it to BA [10 metres] and use TRIGONOMETRIC RATIOS to arrive at our conclusion. Just in case you have forgotten what they were, here they are:

$\displaystyle \frac{OPPOSITE}{HYPOTENUSE} = sin\:θ \\ \frac{ADJACENT}{HYPOTENUSE} = cos\:θ \\ \frac{OPPOSITE}{ADJACENT} = tan\:θ \\ \frac{HYPOTENUSE}{ADJACENT} = sec\:θ \\ \frac{HYPOTENUSE}{OPPOSITE} = csc\:θ \\ \frac{ADJACENT}{OPPOSITE} = cot\:θ$

We can now solve for EC:

$\displaystyle cot\:60 = \frac{12}{EC} → ECcot\:60 = 12 \\ \\ \frac{ECcot\:60}{cot\:60} = \frac{12}{cot\:60} → 12\sqrt{3} = EC$

OR

$\displaystyle tan\:60 = \frac{EC}{12} → 12tan\:60 = EC → 12\sqrt{3} = EC$

Now that you have solved for EC, you can now add it to your original 10 metres to get $\displaystyle 10 + 12\sqrt{3}$ metres. As a decimal, you would get $\displaystyle 30,78460969...$ metres. You can go ahead and round this off if necessary.

** The reason why the cotangent [or tangent] ratio was used was because EA is equivalent to DB by the definition of a rectangle. It has two pairs of parallel and congruent sides with four right angles. Plus, that is the adjacent side of the triangle, while EC is the opposite side of the triangle, so we knew our ratios were correct.

I am joyous to assist you at any time.