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andreyandreev [35.5K]
2 years ago
9

What error did Maria likely make?

Mathematics
1 answer:
raketka [301]2 years ago
7 0
She didn’t notice that the 3 and the 4 both do not have the variable “n” with them
She just added up all the numbers together then added the variable at the end.
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1. Write a user-defined Matlab function for the following math function: y(x) = (-0.2x3 + 7x2)e-0.3x
erica [24]

Answer:B

Step-by-step explanation: the answer is B

3 0
2 years ago
Arc CD is One-fourth of the circumference of a circle. What is the radian measure of the central angle?
Alex

Answer:

\frac{\pi }{2}

Step-by-step explanation:

Since the arc is \frac{1}{4} of the circumference, then

central angle = \frac{1}{4} × 2π ← angle in whole circle

central angle = \frac{2\pi }{4} = \frac{\pi }{2}

8 0
2 years ago
Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be
Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of t=0.6482

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

h(t)=-16t^2 +v_{o}t+s       Eqn(1).

Where:

h(t) : is the height range as a function of time

v_{o}   : is the initial velocity

s     : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

h(t)=-16t^2 + v_{o}t + 40        Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

<u>Ben:</u>

We know that v_{o}=60ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+60t+40        Eqn.(3)

<u>Josh:</u>

We know that v_{o}=48ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+48t+40       Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em>h(t)=0<em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

Thus we have for Ben, Eqn. (3) gives:

h(t)=0-16t^2+60t+40=0

Using the quadratic expression to find the roots of the quadratic we have:

t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec

Since time can only be positive we reject the t_{2} solution and we keep that Ben's book took t=4.3276 seconds to reach the ground.

Similarly solving for Josh we obtain

t_{1}=3.6794sec\\t_{2}=-0.6794sec

Thus again we reject the negative and keep the positive solution, so Josh's book took t=3.6794 seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482

So to answer the original question:

<em>Whose textbook reaches the ground first and by how many seconds?</em>

  • Josh's textbook reached the ground first
  • Josh's textbook reached the ground first by a difference of t=0.6482

3 0
2 years ago
A ship anchored in a port has a ladder which hangs over the side. The length of the ladder is 200cm, the distance between each r
bulgar [2K]

Answer:

The answer to your question is 8 h

Step-by-step explanation:

Data

length of the ladder = 200 cm

distance between each rung = 20 cm

rate = 10 cm/h

fifth rung = ?

Process

1.- Calculate the total distance the tide must rise

distance = 20 cm x 4

              = 80 cm              because the first rung touches the water

2.- Calculate the time

rate = distance / time

-Solve for time

time = distance / rate

-Substitution

time = 80 cm / 10cm/h

-result

time = 8 h

3 0
1 year ago
How would u find 4 1\5 divided by 2 1\3
Over [174]
Remmber
(a/b)/(c/d)=(a/b)(d/c)=(ad)/(bc)
conver to improper

4 and 1/5=20/5+1/5=21/5
2 and 1/3=6/3+1/3=7/3

(21/5)/(7/3)=(21/5)(3/7)=63/35=9/5=1 and 4/5
3 0
1 year ago
Read 2 more answers
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