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3 years ago

Does y = 22x - 65 represent a linear relationship

Mathematics

1 answer:

Luba_88 [7]3 years ago

5 0

yes is a linear line because linear mean a straight line

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An isosceles right triangle has sides that are x+2 units long and a hypotenuse that is 8 units long. What is the length of the m

blsea [12.9K]

Answer:

In an isosceles right triangle, the hypotenuse is larger than the sides by a factor of the square root of 2.

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3 years ago

Please and thank you help me

Art [367]

9514 1404 393

Answer:

your graph is correct
y = 3x -5

Step-by-step explanation:

The graph shows you the line has a "rise" of 1-(-2) = 3 for a "run" of 2-1 = 1, so the slope is 3/1 = 3.

The y-intercept is seen to be -5, so the equation is ...

y = mx +b . . . . . where m is the slope (3) and b is the y-intercept (-5)

y = 3x -5

The points and line are plotted correctly, so you already have your graph.

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3 years ago

A firework is launched at the rate of 10 feet per second from a point on the ground 50 feet from an observer. to 2 decimal place

Kazeer [188]

The rate of change of the angle of elevation when the firework is 40 feet above the ground is 0.12 radians/second.

First we will draw a right angle triangle ΔABC, where ∠B = 90°

Lets, assume the height(AB) = h and base(BC)= x

If the angle of elevation, ∠ACB = α, then

tan(α) = $\frac{AB}{BC} = \frac{h}{x}$

Taking inverse trigonometric function, α = tan⁻¹ ( $\frac{h}{x}$ ) .............(1)

As we need to find the rate of change of the angle of elevation, so we will differentiate both sides of equation (1) with respect to time (t) :

$\frac{d\alpha}{dt}=[\frac{1}{1+ \frac{h^2}{x^2}}]*(\frac{1}{x})\frac{dh}{dt}$

Here, the firework is launched from point B at the rate of 10 feet/second and when it is 40 feet above the ground it reaches point A,

that means h = 40 feet and $\frac{dh}{dt}$ = 10 feet/second.

C is the observer's position which is 50 feet away from the point B, so x = 50 feet.

$\frac{d\alpha}{dt}= [\frac{1}{1+ \frac{40^2}{50^2}}] *\frac{1}{50} *10\\ \\ \frac{d\alpha}{dt} = [\frac{1}{1+\frac{16}{25}}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt} = [\frac{25}{41}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt}= \frac{5}{41} =0.1219512$

= 0.12 (Rounding up to two decimal places)

So, the rate of change of the angle of elevation is 0.12 radians/second.

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3 years ago