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aliina [53]
3 years ago
9

The debate team is showing a video of their recent debate. The first showing begins at 3:15 p.m. The second showing starts at 4:

50 p.m with a 1/2 break between the showings. The second showing started 20 minutes late. Will the second showing be over by 6:30 p.m? EXPLAIN you answer.
Mathematics
1 answer:
Ghella [55]3 years ago
7 0
Given:
3:15 pm - start of first showing.
30 minutes break;
20 minutes late - 2nd showing.
4:50 - start of second showing.

4:50 - 0:20 = 4:30 should be the start of the second showing.
4:00 - 4:30 = 1/2 hour break
3:15 - 4:00 = duration of the 1st showing. 45 minutes in all.

4:50 + 0:45 = 5:35 pm end of the 2nd showing.

Yes, the 2nd showing will be over by 6:30 pm. 
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The perimeter of a rectangle is 200 meters. If the width is 25% of the length, find the dimensions of the rectangle
Eva8 [605]

Answer:

The rectangle dimensions are 20 m by 80 m

Step-by-step explanation:

Let W represent the width and L the length.  Then W = 0.25L.

The perimeter, P, is 200 m.  The formula for that is P = 2W + 2L.

Substituting 0.25L for W, we get:

2(0.25L) + 2L = 200 m, or (after mult. both sides by 2):  L + 4L = 400 m

Then 5L = 400 m, and L = 80 m.  Thus, W = 0.25(80 m) = 20 m.

The rectangle dimensions are 20 m by 80 m

8 0
3 years ago
There are 181 students going on a college field trip.
Norma-Jean [14]

Answer:

42.5 (students that are on each bus)

Step-by-step explanation:

181 (students)-11 (students in cars) =170 students left

170(students left)÷4(buses)= 42.5(students that are on each bus)

Hope this help you

3 0
3 years ago
Help ASAP! A person invests 10000 dollars in a bank. The bank pays 4.5% interest compounded
Roman55 [17]

Answer: is a=p(1add7)by

A=8pnt,

Step-by-step explanation:

3 0
3 years ago
NASA launches a rocket at t=0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t)=−4.9t2+121t
Alenkasestr [34]

NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by

h(t)=-4.9t^2+121t+283

The sea level is represented by h = 0, therefore, to find the corresponding time when h splashes into the ocean we have to solve for t the following equation:

h(t)=0\implies-4.9t^2+121t+283=0

Using the quadratic formula, the solution for our problem is

\begin{gathered} t=\frac{-(121)\pm\sqrt{(121)^2-4(-4.9)(283)}}{2(-4.9)} \\ =\frac{121\pm142.083778...}{9.8} \\ =\frac{121+142.083778...}{9.8} \\ =26.8452834694... \end{gathered}

The rocket splashes after 26.845 seconds.

The maximum of this function happens at the root of the derivative. Differentiating our function, we have

h^{\prime}(t)=-9.8t+121

The root is

-9.8t+121=0\implies t=\frac{121}{9.8}=12.347

Then, the maximum height is

h(12.347)=1029.99

1029.99 meters above sea level.

6 0
1 year ago
Someone help me with this math PlZZ!!!imma use 80 points!!
LiRa [457]

Answer:

Will you try to please repost it so i can better answer it and try to reword it some because it's a bit confusing for those who want to help me

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3 years ago
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