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3 years ago

Six times a number is greater than 20 more than that number. What are the possible values of that number?

Mathematics

2 answers:

jek_recluse [69]3 years ago

6 0

6n>20+n
6n-n>20
5n>20
n>4

Margarita [4]3 years ago

4 0

Answer: n>4

Step-by-step explanation:

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A plane flying horizontally at an altitude of 3 miles and a speed of 500 mi/h passes directly over a radar station. Find the rat

konstantin123 [22]

Answer:

The rate at which the distance from the plane to the station is increasing is 331 miles per hour.

Step-by-step explanation:

We can find the rate at which the distance from the plane to the station is increasing by imaging the formation of a right triangle with the following dimensions:

a: is one side of the triangle = altitude of the plane = 3 miles

b: is the other side of the triangle = the distance traveled by the plane when it is 4 miles away from the station and an altitude of 3 miles

h: is the hypotenuse of the triangle = distance between the plane and the station = 4 miles

First, we need to find b:

$a^{2} + b^{2} = h^{2}$ (1)

$b = \sqrt{h^{2} - a^{2}} = \sqrt{(4 mi)^{2} - (3 mi)^{2}} = \sqrt{7} miles$

Now, to find the rate we need to find the derivative of equation (1) with respect to time:

$\frac{d}{dt}(a^{2}) + \frac{d}{dt}(b^{2}) = \frac{d}{dt}(h^{2})$

$2a\frac{da}{dt} + 2b\frac{db}{dt} = 2h\frac{dh}{dt}$

Since "da/dt" is constant (the altitude of the plane does not change with time), we have:

$0 + 2b\frac{db}{dt} = 2h\frac{dh}{dt}$

And knowing that the plane is moving at a speed of 500 mi/h (db/dt):

$\sqrt{7} mi*500 mi/h = 4 mi*\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{\sqrt{7} mi*500 mi/h}{4 mi} = 331 mi/h$

Therefore, the rate at which the distance from the plane to the station is increasing is 331 miles per hour.

I hope it helps you!

4 0

3 years ago

Solve the compound inequality. 6 – x > 15 or 2x – 9 ≥ 3 A. x > 9 or x ≤ 6 B. x < 9 or x ≥ –6 C. x > –9 or x ≤ –6 D.

Vesnalui [34]

Answer:

Step-by-step explanation:

Solve the compound inequality. 6 – x > 15 or 2x – 9 ≥ 3 A. x > 9 or x ≤ 6 B. x < 9 or x ≥ –6 C. x > –9 or x ≤ –6 D. x < –9 or x ≥ 6

4 0

3 years ago

1 cup = ______ fluid ounces = 8

Brut [27]

Answer:

Step-by-step explanation:

5 0

3 years ago

If g(r) =2(r-1), Find g(10)

garri49 [273]

Answer:

g(r) = 18

Step-by-step explanation:

Substitute 10 into r

g(10) = 2( (10) -1)

= 2(9)

= 18

7 0

3 years ago

If two objects travel through space along two different curves, it’s often important to know whether they will collide. (Will a

galina1969 [7]

Answer:

Both objects collide at t=3 in the point <9,9,9>

Step-by-step explanation:

Collision Of Moving Objects

Two objects can describe different trajectories in the space. Those trajectories can intersect in one or more points but it doesn't mean they collide. Collision occurs if they are in the same position at the same time. If we know the positions as a function of time of each object, we could try so find if, for a given time, they are in the same position.

The positions of two object are given as

$r1(t)=$

$r2(t)=$

Let's find out if there is at least one value of t that makes both positions to be the same. We can try by equating one of the three coordinates and testing if the value of t make both have the same x,y,z coordinate. Let's try equating the x-components of both

$t^2=4t-3$