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3 years ago

7

At noon, the shadow of a flagpole is 10 feet long. At the same time, the shadow of a 140-foot tall building is 60 feet long. Wha

t is the height of the flagpole? Round your answer to the nearest foot.

1 answer:

Free_Kalibri [48]3 years ago

6 0

10=2x5
60=2x2x3x5

2x2x5x3=60

i think to do this you need lcm but im not sure

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Today, 6 friends went out for lunch. Their total bill was $50.41 , including tax and gratuity. One of them ordered only soda and

Burka [1]

Answer:

50.41 minus 1.71 is 48.70 divided by 5 is 9.74

3 0

3 years ago

Answer the question below. Be sure to show your work

likoan [24]

ANSWER:

We have to find new side-lengths of PQR triangle.

Original sides are.

$\begin{gathered} PQ\text{ = 8cm} \\ QR=17\operatorname{cm} \\ RP=15\operatorname{cm} \end{gathered}$

After multiplying by 2.5 we get

$\begin{gathered} PQ^{\prime}=2.5\times PQ=(8\times2.5)cm=20\operatorname{cm} \\ QR^{\prime}=(17\times2.5)cm=42.5\operatorname{cm} \\ RP^{\prime}=(15\times2.5)cm=37.5\operatorname{cm} \end{gathered}$

These are the new sides of the P'Q'R' triangle.

5 0

1 year ago

The University of Montana ski team has thirteen entrants in a men's downhill ski event. The coach would like the first, second,

sleet_krkn [62]

Answer:

1716 ways

Step-by-step explanation:

Given that :

Number of entrants = 13

The number of ways of attaining first, second and third position :

The number of ways of attaining first ; only 1 person can be first ;

Using permutation :

nPr = n! ÷(n-r)!

13P1 = 13! ÷ 12! = 13

Second position :

We have 12 entrants left :

nPr = n! ÷(n-r)!

12P1 = 12! ÷ 11! = 12

Third position :

We have 11 entrants left :

nPr = n! ÷(n-r)!

11P1 = 11! ÷ 10! = 11

Hence, Number of ways = (13 * 12 * 11) = 1716 ways

8 0

3 years ago

Ill give you alot of points for this

rusak2 [61]

Answer:

ok

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

The sum of two integers is 33. The larger is 6 more than twice the smaller. Find the two

grandymaker [24]

Answer:

24 and 9

Step-by-step explanation:

Hi there!

Let x be equal to the larger integer.

Let y be equal to the smaller integer.

<u>1) Construct equations</u>

$x+y=33$ (The sum of two integers is 33)
$x=6+2y$ (The larger is 6 more than twice the smaller)

<u>2) Solve for one of the integer</u>

Isolate x in the first equation

$x+y=33\\x=33-y$

Plug the first equation into the second

$33-y=6+2y$

Combine like terms

$33-6=2y+y\\27=3y\\9=y$

Therefore, the smaller integer is 9.

<u />

<u>3) Solve for the other integer</u>

$x+y=33$

Plug in y (9)

$x+9=33\\x=33-9\\x=24$

Therefore, the larger integer is 24.

I hope this helps!

4 0

3 years ago

Read 2 more answers