How do you graph it on a line

Mathematics

1 answer:

Aleonysh [2.5K]3 years ago

8 0

In order to draw x ≤ -10 on a number line, first draw a closed dot at -10. This is because we have a ≤ and not a <. If we had a <, it would be an open dot.

Next, since we want x to be smaller, draw an arrow pointing towards all numbers that are smaller. This arrow would point to the left.

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Need help asap please...

son4ous [18]

Answer: Option 'B' is correct.

Step-by-step explanation:

Since we know that "Angle opposite to greater side is smaller or angle opposite to smaller side is larger".

So, ∠AOD is opposite to side 'AD' which is equal to 8.11

Similarly, ∠AOB is opposite to side 'AB' which is equal to 8.

So, ∠AOB>∠AOD

Hence, Option 'B' is correct.

3 0

3 years ago

What is the solution to -2(4x + 8) 4(2 - x)?

Gnesinka [82]

Answer:b

Step-by-step explanation:

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7 0

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.

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