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2 years ago

2/5x - 4 = 10 solve for x

Mathematics

1 answer:

lawyer [7]2 years ago

5 0

Answer: x = +35

Explanation: In this equation, our first step is to isolate the x-term by adding 4 to both sides of the equation. On the left -4 and +4 cancel and we're left with 2/5x and on the right 10 + 4 simplifies to 14.

Now, in order to get x by itself, we must multiply both sides of the equation by the reciprocal of that fraction. The reciprocal of a fraction is just that fraction flipped so the reciprocal of 2/5 is 5/2 so we multiply both sides of the equation by 5/2.

So as we multiply 5/2 times 5/2x, our 5's cancel and our 2's cancel so we are simply left with x. On the right side, 14 and 2 reduce to 7 and 1 so we have 7 times 5 which is 35 so x = 35.

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kotegsom [21]

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Multiply by 6xy to clear the fractions.

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That's a second degree equation, also known as a conic. That one happens to be a hyperbola.

$30x = y(13 x +12)$

$y = \dfrac{30x}{13 x + 12}$

Let's clear the fractions from the second equation, multiplying out common denominator xy:

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$36y - 24 x = xy$

We are being asked to find the meet of two hyperbolas, so we expect two answers, a quadratic equation.

Substituting,

$36 \left( \dfrac{30x}{13 x + 12} \right) - 24 x = x \left( \dfrac{30x}{13 x + 12} \right)$

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A surveyor standing on top of a cliff found that the angle of depression of a boat as sea is 22°. if the top of the cliff is 15m

Alex73 [517]

Answer:

From the photo above i interpreted the question and drew a triangle to make it easier to solve

Tan is used to solve the question as we were given opposite (as height of the cliff) and to find Adjacent ( the distance between the foot of the cliff to the boat)

Tan θ =Opp/Adj

Tan 22°=15/a

Cross Multiply

aTanθ=15°

Divide both sides by Tan 22

aTan22/Tan22 = 15/Tan 22

a = 15/Tan 22°

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Therefore the distance between the foot of the cliff to the boat is 37.13 metres

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2 years ago

20 points!

Arte-miy333 [17]

Answer:

Step-by-step explanation:

hypotenuse=2*AB

where AB is the smallest side.

Take AB as the base .

Draw a perpendicular at B.

A as the center and cut an arc AC=2 AB

Join AC.

CAB is the reqd. triangle.

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