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Reil [10]
3 years ago
6

How do I work this out???

Mathematics
1 answer:
Finger [1]3 years ago
7 0
Okay so form a right angle triangle and the angle of that is 65 as it is right angle so 90 - 25= 65. label the sides like the pic below. 

we have the hypothenuse as well as the angle and we are trying to find the o(opposite side) 

o/h =sin
sin 65=23/h
sin 65 = 23/x
23/sin 65=x 
x=25.4
how high the kite is above ground is :
= 25.4 +1.2
 =26.6

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What is the inverse function of f(x) = 25x-3?
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Answer:

The answer is

<h2>{f}^{ - 1} (x) =  \frac{x +3 }{25}</h2>

Step-by-step explanation:

f(x) = 25x - 3

To find the inverse of f(x) , equate f(x) to y

That's

y = f(x)

We have

y = 25x - 3

Next interchange the terms that's x becomes y and y becomes x

x = 25y - 3

<u>Next solve for y</u>

Move 3 to the other side of the equation

That's

25y = x +  3

<u>Divide both sides by 25</u>

\frac{25y}{25}  =  \frac{x + 3}{25}

We have the final answer as

{f}^{ - 1} (x) =  \frac{x +3 }{25}

Hope this helps you

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\\ \sf\longmapsto 2x+y=2

\\ \sf\longmapsto 2x=2-y

\\ \sf\longmapsto x=\dfrac{2-y}{2}\dots(1)

And

\\ \sf\longmapsto x+1=y+2

\\ \sf\longmapsto x=y+2-1

\\ \sf\longmapsto x=y+1

  • Put the value

\\ \sf\longmapsto \dfrac{2-y}{2}=y+1

\\ \sf\longmapsto 2-y=2(y+1)

\\ \sf\longmapsto 2-y=2y+2

\\ \sf\longmapsto 2-2=2y+y

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\\ \sf\longmapsto y=\dfrac{0}{3}

\\ \sf\longmapsto y=\infty

  • Put in eq(1)

\\ \sf\longmapsto x=\dfrac{2-y}{2}

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5 0
3 years ago
Read 2 more answers
The temperature at a point (x, y) is T(x, y), measured in degrees Celsius. A bug crawls so that its position after t seconds is
aniked [119]

Answer:

10.50°C

Step-by-step explanation:

Given x = 2 + t , y = 1 + 1/2t where x and y are measured in centimeters. Also, the temperature function satisfies Tx(2, 2) = 9 and Ty(2, 2) = 3

The rate of change in temperature of the bug path can be expressed using the composite formula:

dT/dt = Tx(dx/dt) + Ty(dy/dt)

If x = 2+t; dx/dt = 1

If y = 1+12t; dy/dt = 1/2

Substituting the parameters gotten into dT/dt we will have;

dT/dt = 9(1)+3(1/2)

dT/dt = 9+1.5

dT/dt = 10.50°C/s

Hence the rate at which the temperature is rising along the bug's path is 10.50°C/s

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