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3 years ago

How many groups are of sixty are in two hundred forty four

Mathematics

2 answers:

Semenov [28]3 years ago

6 0

244/60=4.06666666

So 4 groups

Fudgin [204]3 years ago

6 0

Simply divide 244 by 60.

244 / 60 = 4.066 repeating, which we can round down to 4.

There are 4 groups of 60 in 244.

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Anna11 [10]

140 degrees because an angle of a circle is 360 degrees and you can subtract these angles from it to get 140 degrees

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3 years ago

Read 2 more answers

Each morning Rin rides 1.5 miles to school and then rides home in the afternoon. Later in the evening, she rides to the park and

brilliants [131]

Answer:

Last equation given in the list of possible answers:

5 ( 1.5 + 1.5 + x ) = 25

Step-by-step explanation:

We need to include in the total addition of miles ridden during the week:

a) 1.5 miles to the school

b) 1.5 miles from school back home

c) x miles for the evening ride

so for the miles ridden per day we have: "1.5 +1.5 + x"

Now, since per week she does 5 days like this, then we need to multiply the expression above by 5 in order to total the number of miles she rides weekly (25 miles)

5 ( 1.5 + 1.5 + x ) = 25

And we can use this equation to find the amount "x" that Rin rides in the evening.

7 0

3 years ago

If a/b < c/d with b > 0, d > 0, prove that a+c / b+d lies between the two fractions a/b and c/d .

Tems11 [23]

Divide through everything by b :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}$

Since a/b < c/d, it follows that

$\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}$

Multiply through everything on the right side by b/d to get

$\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd$

and so (a + c)/(b + d) < c/d.

For the other side, you can do something similar and divide through everything by d :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}$

and a/b < c/d tells us that

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}$

Then

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ab + \dfrac{ad}{b^2}}{1 + \dfrac db} = \dfrac{\dfrac ab\left(1+\dfrac db\right)}{1 + \dfrac db} = \dfrac ab$

and so (a + c)/(b + d) > a/b.

Then together we get the desired inequality.

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3 years ago

If x,y and z are positive integers and 3x=4y=7z,

Pani-rosa [81]

3x=4y=7z
so
GCF:
3 * 4 * 7 = 84

3x = 84
x = 28

4y = 84
y = 21

7z = 84
z = 12

least possible value for
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answer
(d) 61

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3 years ago

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borishaifa [10]

Answer:

12.2 12.3 12.4 12.5

Step-by-step explanation:

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