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sergiy2304 [10]
3 years ago
13

(1). An Aeroplane leaves an airport, flies due North for 2 hrs at 500km/h. It then flies on a bearing of 053 degree at 300km/h f

or 90 mins. How far is the plane from the airport correct to 1 dp *(2). Find the bearing of the plane from the Airport correct to 1 dp. *(3). If it takes the Aeroplane 150 mins to fly back to the airport. Find the average speed of the aeroplane for the whole flight correct to 1dp. *

Mathematics
1 answer:
Agata [3.3K]3 years ago
4 0

Answer:

1. 1320.7 km

2. 015.8 degrees.

3. 461.8 km/h.

Step-by-step explanation:

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What is 3/4 - 5/7 ?
AveGali [126]

The solution of \frac{3}{4} - \frac{5}{7} is \frac{1}{28} \text{ or } 0.0357

<em><u>Solution:</u></em>

Given that we have to find the solution of \frac{3}{4} - \frac{5}{7}

To solve the given sum, first make the denominators of both the fractions same

This can be done by taking L.C.M of both denominators

<em><u>Step 1:</u></em>

L.C.M of 4 and 7:

The prime factor of 4 = 2 x 2

The prime factor of 7 = 7

For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.

The new superset list is

2, 2, 7

Multiply these factors together to find the LCM

LCM = 2 x 2 x 7 = 28

Thus L.C.M of denominators is 28

<em><u>Step 2:</u></em>

Solution of \frac{3}{4} - \frac{5}{7}

Multiply the denominator by a number to get 28 and multiply that same number with numerator also

\frac{3}{4} - \frac{5}{7} = \frac{3 \times 7}{4 \times 7} - \frac{5 \times 4}{7 \times 4}=\frac{21}{28} - \frac{20}{28}\\\\\frac{21}{28} - \frac{20}{28} = \frac{21-20}{28} = \frac{1}{28}\\\\\frac{1}{28} = 0.0357

Thus solution of \frac{3}{4} - \frac{5}{7} is \frac{1}{28} \text{ or } 0.0357

3 0
3 years ago
Solve the proportion. Please explain how you managed to get the answer! I don't know how to get there!
belka [17]
All you need to do is cross multiply and then solve for x
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3 0
3 years ago
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Soloha48 [4]

Answer:

Step-by-step explanation:

25- guaranteed

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238+25= $263

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2. He expects to make 25+14t per day.

5 0
2 years ago
Simplify the expression: (− 2/3 pq^4)^2·(−27p^5q)
VladimirAG [237]

Answer:

-12p^7q^9

Step-by-step explanation:

(-2/3Pq^4)^2×(-27P^5q)

= ((-2)^2)/3^2 × P^2 q^4-2 × (-27qp^5)

= ((-2)^2)/3^2 × 27p^7q9

= (((-2)^2 x 27p^7q9)/3^2

= -((4×27p^7q^9)/9)

= -(108p^7q^9)/9

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4 0
3 years ago
The tangent to the circle x^2+y^2=18 is parallel to the tangent x+y=6
liubo4ka [24]

Answer:

y = -x - 6

Step-by-step explanation:

If we are looking for the equation of the line tangent to the circle, we have to find the first derivative of the circle function.  If that tangent line is to be parallel to the given tangent line, we need to find the slope of the the given tangent line and make sure the slope of the line we are looking for is the same.  First let's find the slope of the given tangent.  If the given tangent is x+y=6, then to find the slope, solve for y:

y = -x + 6.  So the slope of that line, and also the slope of the tangent we are solving for, is -1.  Hold that thought while we find the derivative of the function.  Using implicit differentiation, we find the derivative to be:

2x+2y\frac{dy}{dx}=0

Solving for the slope (dy/dx) gives us:

2y\frac{dy}{dx}=-2x and

\frac{dy}{dx}=\frac{-2x}{2y} so

\frac{dy}{dx}=-\frac{x}{y}

Subbing in the value of the slope we found:

-1=-\frac{x}{y} so

-y = -x or, equivalently,

y = x.  Now that we know that y and x are the same value, we can go back to the original circle to find out what they both are by substitution.  If y = x, we make the substution:

If x^2+y^2=18, then

y^2+y^2=18 and

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y = ±3

We will choose the negative root (you'll see why in a second) and sub that in for both y and x since y and x are th same number.  Going back to the fact that the slope is -1:

y - (-3) = -1(x - (-3)) simplifies a bit to:

y + 3 = -x - 3 which gives us, in slope-intercept form:

y = -x - 6

That is the equation of the tangent to the circle that is parallel to the given tangent.

If we would have chosen the principle (or positive) root of 3, our equation would have looked like this:

y - 3 = -1(x - 3) and

y = -x + 3 + 3 so

y = -x + 6.  Notice that that is the EXACT SAME EQUATION as the given tangent.  That's why we have to pick the negative 3 as our root.

Good luck in your calculus class!  Make sure you post your questions in here!  Many of us LOVE the challenge of calculus!

3 0
3 years ago
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