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lisov135 [29]
3 years ago
10

If movie starts at 2:30 and is 112 minutes, what time will the movie end?

Mathematics
2 answers:
notsponge [240]3 years ago
7 0

At the hour of 4:22 just add

Oduvanchick [21]3 years ago
7 0
2:30- 3:30 60 min

3:30-4:00 30 min

4:00- 4:22 22 min

the movie ends at 4:22
You might be interested in
A circle has radius 50 cm. Find the area of the circle to the nearest hundredth.
padilas [110]
Multiply by 2 which will get you a diameter and then multiply it by pi which is 3.14.
7 0
2 years ago
There are 800 desks at Washington Middle School. The custodial staff at the school plans to inspect 6.5% of the desks every week
Talja [164]

Answer:

208 desks

Step-by-step explanation:

To find out how many desks they will inspect each week, we multiply

800 * 6.5%

800 * .065

52 desks per week

Over a 4 week period, they will inspect

52 desk per week * 4 weeks =

208 desks

7 0
3 years ago
The question is what is other equation to find 16-9
disa [49]
Its 9-1 so that would be 8 idk if you are right just noticed and though it would help 
4 0
3 years ago
I need help before tomorrow
Hitman42 [59]

Hello from MrBillDoesMath!

Answer:

The equations "are balanced"


Discussion:

Given that

4 + 4 = 8                 => subtracting k from both sides

4 + 4 - k = 8 -k


so the equation shown IS correct. The answer should be the equations "are balanced"


Thank you,

MrB

8 0
3 years ago
Which is the simplified form of the expression ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript nega
AlekseyPX

Answer:

The option "StartFraction 1 Over 3 Superscript 8" is correct

That is \frac{1}{3^8} is correct answer

Therefore [(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}

Step-by-step explanation:

Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared

The given expression can be written as

[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2

To find the simplified form of the given expression :

[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2

=(2^{-2})^{-3}(3^4)^{-3}\times (2^{-3})^2(3^2)^2 ( using the property (ab)^m=a^m.b^m )

=(2^6)(3^{-12})\times (2^{-6})(3^4) ( using the property (a^m)^n=a^{mn}

=(2^6)(2^{-6})(3^{-12})(3^4) ( combining the like powers )

=2^{6-6}3^{-12+4} ( using the property a^m.a^n=a^{m+n} )

=2^03^{-8}

=\frac{1}{3^8} ( using the property a^{-m}=\frac{1}{a^m} )

Therefore [(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}

Therefore option "StartFraction 1 Over 3 Superscript 8" is correct

That is \frac{1}{3^8} is correct answer

6 0
3 years ago
Read 2 more answers
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