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Troyanec [42]
3 years ago
7

the measure of angle a is 16 degrees greater than the measure of angle b. the tow angles are complementary. find the measure of

each angle
Mathematics
1 answer:
Fudgin [204]3 years ago
4 0
This is a system of equations problem.  Set up the 2 equations like so: If the angles are complementary then they add up to 90, therefore, a + b = 90.  We also know that a is 16 more than b.  The word "is" means equals and "more" is addition.  Therefore, a is 16 more than b is "a = b + 16".  Now sub in that value of a (b + 16) into the first equation and solve for b.  Then back-substitute to solve for a. (b + 16) + b = 90 so 2b + 16 =  90 and 2b = 74.  So b = 37.  If b = 37, then a + 37 = 90 and a = 53.  Check yourself to make sure that 37 + 53 add up to equal 90 (they do, just get used to checking yourself for accuracy).

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Step-by-step explanation:

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A 10p coin weighs 6.5g. What is the weight of a bag of 10p coins worth £3.50?​
hichkok12 [17]

Answer:

227.5 grams

Step-by-step explanation:

We know that a single 10p coin weights 6.5 g

Now we want to find the weight of a bag of 10p coins, such that the net value is £3.50  (where the weight of the bag is neglected)

The value of a 10p coin is £0.10

So the first thing we need to find, is how many coins there are in the bag.

To find that, we need to find the quotient between the total value and the value of a single coin:

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7 0
3 years ago
A floppy disc can store 1440000 bytes of data
max2010maxim [7]

Answer:

1440000 = 1.44\times 10^{6}

We need approximately 1667 floppy disc to store 2.4\times 10^9 bytes of data.      

Step-by-step explanation:

We are given the following information in the question:

A floppy disc can store 1440000 bytes of data.

We have to convert  1440000 in a standard form.      

Standard Form:

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The standard form can be written as:

1440000 = 1.44\times 10^{6}

We have positive power of 10 because we have to move to the right of the decimal point.

A hard disc can store 2.4\times 10^9 bytes of data.

Number of floppy disks needed to store the 2.4\times 10^9 bytes of data =

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