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2 years ago

A small telescope magnifies

Mathematics

1 answer:

WARRIOR [948]2 years ago

4 0

The observatory telescope has 22 times the magnifying power compared to the small telescope.

Data;

small telescope = 150 times
large telescope = 3300

<h3>Division of Numbers</h3>

To solve this question, we have to divide the large telescope by the small telescope to calculate it's magnifying power.

$x = \frac{3300}{150}=22$

The magnification of the observatory telescope is 22 times greater or larger than the small telescope.

Learn more on division of numbers here;

brainly.com/question/7068223

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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.

Ben:

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

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

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

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

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 $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

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:

Whose textbook reaches the ground first and by how many seconds?

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

3 0

3 years ago

Do you know What is X=9+y-2

Damm [24]

Answer:

Subtract

from

Step-by-step explanation:

5 0

3 years ago

Can anybody help? I do not understand:(

Dvinal [7]

Six now have a good day happy face bye-bye

7 0

3 years ago

the human heart pumps 750 gallons of blood in 9 hours. A human kidney filters 100 gallons of blood in 6 hours. How many more gal

WITCHER [35]

The heart pumps 750*24/9=2000 gallons, and the kidney pumps 100*24/6=400 gallons. This is a difference of 2000-400=1600 gallons.

3 0

3 years ago

Which number will have the same result when rounded to the nearest ten or hundred A-97 B-118and5 C-179 D-5091

slamgirl [31]

Lets round it to the nearest ten
A 97 ====> 100
B 118 ===> 120
C 179 ===> 180
D 5091 ==> 5090
No result yet, lets round to the nearest hindred.
A 97 ====> 100
B 118 ===> 100
C 179 ===> 180
D 5091 ==> 5100
As we can see only A give the same result when we round it to the nearest hundred and nearest ten.

5 0

4 years ago