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dalvyx [7]
1 year ago
11

Bob and joe have a class together at Kensington college. They leave the class at the same time. Bob goes to the library at 4 mil

es per hour, joe heads to the exact opposite direction at 2 miles per hour. How long will it be before the classmates are 2 miles apart .
Mathematics
1 answer:
galben [10]1 year ago
5 0

Solution

For this case we can create the following schema

We can use the definition of distance:

x= vt

We can find the relative velocity:

V_{\text{rel}}=4-2=2\frac{mi}{hr}

And the distance between both is required to be 2 mi so then we can find the time like this:

t=\frac{x}{v_{\text{rel}}}=\frac{2mi}{\frac{2mi}{hr}}=1hr

Then the answer is:

1 hr

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Answer:

is this a joke? xd. 2+2=4.

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Michael added 15 fish to his pond over a period of 5 days he added the same number of fish each day what was the change in the n
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5 0
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Gekata [30.6K]
Hello there.

<span>Which line is a graph of the equation 2x + 5y = 10?

Coordinate plane with the following lines graphed: Line A contains a slope of five-halves and y-intercept of 2; Line B contains a slope of negative five-halves and y-intercept of 2; Line C contains a slope of five-halves and y-intercept of negative 2; and Line D contains a slope of negative five-halves and y-intercept of negative 2. 

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7 0
3 years ago
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Right triangle ABC is similar to right triangle DEF. If the side lengths for triangle ABC are 15, 20, 25, respectively, which va
solniwko [45]

Answer:

A = 15   B = 20   C = 25

D = 30   E = 40   F = 50

D = 3    E = 4     F = 5

Step-by-step explanation:

Given

Similar Triangles: ABC and DEF

A = 15

B = 20

C = 25

Required

Determine the sides of DEF

No options were given, so I will solve on a general terms.

Since both triangles are similar, then the following relationship exists.

DEF = ABC * n

i.e.

D = A * n    E = B * n     F = C * n

Where

n = Scale\ Factor

Assume n = 2.

So, we have:

D = A * n    

D =15 *2

D = 30

E = B * n    

E = 20 * 2

E = 40

F = C * n

F = 25 * 2

F = 50

Assume n = \frac{1}{5}.

So, we have:

D = A * n    

D =15 *\frac{1}{5}

D = 3

E = B * n    

E = 20 * \frac{1}{5}

E = 4

F = C * n

F = 25 * \frac{1}{5}

F = 5

So, the possible sides are:

A = 15   B = 20   C = 25

D = 30   E = 40   F = 50

D = 3    E = 4     F = 5

6 0
3 years ago
Can someone plz help me! I’ve been trying to figure it but I just can’t
VMariaS [17]

Answer:

Step-by-step explanation:

I'll walk you through how to get the answers, but I'm not sure what your drop downs there are asking for.  Maybe once you see the answers, you can figure out how to get the drop downs chosen correctly.

For your quadratic function, the h on the far left is how high the ball is after it is kicked and some amount of time has gone by, the -16t-squared part is the pull of gravity in feet per sec per sec, the 55t is the vertical velocity, and the 1 means that the ball was kicked when it was 1 foot off the ground. If we want to find when ("when" is a time measure) the ball was 45 feet off the ground, we sub in 45 for the h on the far left. Remember that that h is how high the ball is off the ground after it is kicked and some time has gone by. When we sub in 45 for that h, we will solve for t by factoring, to see at what times the ball is 45 feet from the ground.

It would be best to set up the equation and then plug it into the quadratic formula. Subbing in 45 for h gives us:

45=-16t^2+55t+1 and we get everything on one side of the equals sign and set the quadratic equal to 0, giving us:

-16t^2+55t-44=0 and then plug it into the quadratic formula with a = -16, b = 55, and c = -44:

t=\frac{-55+/-\sqrt{55^2-4(-16)(-44)} }{2(-16)}  and we'll simplify that a little at a time:

t=\frac{-55+/-\sqrt{3025-2816} }{-32}  and

t=\frac{-55+/-\sqrt{209} }{-32} giving us 2 solutions:

t=\frac{-55+14.45683229}{-32}=1.2669 sec and

t=\frac{-55-14.45683229}{-32}=2.1705 sec

Let's interpret these solutions. When the ball is kicked up into the air, it experiences parabolic travel. It goes up to a certain max height until gravity wins and the ball comes back down to the ground. Everything thrown up into the air experiences parabolic (or projectile) motion (because gravity affects everything!). So the ball with reach a max height of some number of feet (you could find that out if you needed to, but we'll let that one go for now since this isn't physics class!) before it comes back down.  On its way up to that max height, it will pass the 45 foot mark, and when it's coming back down, it will pass it again. The time that it passes the 45 foot mark going up is 1.2669 seconds after it is kicked. Then it goes up as high as it's going to and comes back down, passing the 45 foot mark on the way back down, at 2.1705 seconds after it was kicked.

However you need to use that interpretation to pick the correct drop-down answer is up to you. Much luck with your quadratics!! They're very important; you haven't seen the last of them by far!

5 0
3 years ago
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