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3 years ago

Solve for m . 3m = 9

Mathematics

1 answer:

Rzqust [24]3 years ago

6 0

Answer:

Step-by-step explanation:

9÷3=3 so the answer is 3 and you can check by doing 3×3=9

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Solve for w in P = 2w + 2l, if P = 38 and l = 12.

vova2212 [387]

W= 7

38=2w+2(12)
/
38=2w + 24
-24 -24

14=2w
divid by 2 divid by 2

7=w

4 0

4 years ago

Read 2 more answers

What is the absolute value of |7 1\8|

Studentka2010 [4]

Answer:

Step-by-step explanation:

absolute value of |7 1/8| is 7 1/8

7 0

4 years ago

Two dogs run around a circular track 300 m long. One dog runs at a steady rate of 15m per second, the other at a steady rate of

mihalych1998 [28]

Answer:

300 seconds

Step-by-step explanation:

The first dog run at v₁ = 15 m/sec the second one run at v₂ = 12 m/sec

we know that d = v*t then t = d/v

Then the first dog will take 300/ 12 = 25 seconds to make a turn

The second will take 300 / 15 = 20 seconds to make a turn

Then the first dog in 12 turns 12*25 will be at the start point, and so will the second one at the turn 15.

To check first dog 12 * 25 = 300

And the second dog 15 * 20 = 300

That means that time required for the two dogs to be at the start point together is

300 seconds, in that time the first dog finished the 12 turns, and the second had ended the 15.

Another procedure to solve this problem is as follows:

between 12 m/sec and 15 m/sec the minimum common multiple is 300 ( 300 is the smaller number that accept 12 and 15 as factors 12*15 = 300) Then when time arrives at 300 seconds the two dogs will be again in the starting point

3 0

3 years ago

The angles of depression of two ships from the top of the light house are 45° and 30° towards east. If the ships are 200 m apart

azamat

Answer:

273 meters

Step-by-step explanation:

See image attached for the diagram I used to represent this scenario.

The distance between the ships, at angles 30 and 45, is 200 meters. The distance between the left ship and the lighthouse is x meters.

We can use trigonometric ratios to solve this problem. We can use the tangent ratio $\big{(} \frac{\text{opposite}}{\text{adjacent}} \big{)}$ to create an equation with the two angles.

$\displaystyle \text{tan(45)} = \frac{h}{x}$
$\displaystyle \text{tan(30)} = \frac{h}{x+200} }$

Let's take these two equations and solve for x in both of them.

<h2> $\textbf{Equation I}$ </h2>

$\displaystyle \text{tan(45)} = \frac{h}{x}$

tan(45) = 1, so we can rewrite this equation.

$\displaystyle 1=\frac{h}{x}$

Multiply x to both sides of the equation.

$\displaystyle x = h$

<h2> $\textbf{Equation II}$ </h2>

$\displaystyle \text{tan(30)} = \frac{h}{x+200} }$

Multiply x + 200 to both sides and divide h by tan(30).

$\displaystyle \text{x + 200} = \frac{h}{\text{tan (30)}}$

Subtract 200 from both sides of the equation.

$\displaystyle \text{x} = \frac{h}{\text{tan (30)}} - 200$

Simplify h/tan(30).

$x=\sqrt{3}h - 200$

<h2> $\textbf{Equation I = Equation II}$ </h2>

Take Equation I and Equation II and set them equal to each other.

$h=\sqrt{3}h-200$

Subtract √3 h from both sides of the equation.

$h-\sqrt{3}h=-200$

Factor h from the left side of the equation.

$h(1-\sqrt{3}) =-200$

Divide both sides of the equation by 1 - √3.

$\displaystyle h=\frac{-200}{1-\sqrt{3} }$

Rationalize the denominator by multiplying the numerator and denominator by the conjugate.

$\displaystyle h=\frac{-200}{1-\sqrt{3} } \big{(} \frac{1+\sqrt{3} }{1+\sqrt{3}} \big{)}$
$\displaystyle h=\frac{-200+200\sqrt{3} }{1-3}$
$\displaystyle h =\frac{-200+200\sqrt{3} }{-2}$