Answer:
20
Step-by-step explanation:
<u>step 1: Evaluate the expressions</u>
6*10^7 = 60,000,000
3*10^6 = 3,000,000
<u>step 2: Devide</u>
60,000,000 / 3,000,000 = 20
so 60,000,000 is 20 times larger than 3,000,000
Your teacher might want you to use a different method, but I believe this is the simplest way. If you have any questions feel free to ask away
Answer:
3/8 + 1/4 < 1
Check: Find common denominators, what you do to the denominator, you do to the numerator. Multiply 2 to both the top and bottom of the fraction:
(1/4)(2/2) = 2/8
3/8 + 2/8 < 1
5/8 < 1
5/8 < 8/8 (True).
Any fraction works in the second one, for regardless, you will get less than a whole number, so solve the third one first.
The 2 fractions needed will have to be greater than 1. 1/4 has already been used, so find two pairs of fractions that will, when added, be greater than 1.
(7/8 + 3/4)
Find common denominators, what you do to the denominator, you must do to the numerator:
(7/8)(2/2) + (3/4)(4/4) = (14/16) + (12/16) = 26/16
26/16 > 1 (True).
Finally, the last one will be 2/3.
2/3 - 1/8 < 1
Find common denominators, what you do to the denominator, you must do to the numerator:
(2/3)(8/8) - (1/8)(3/3) < 1
16/24 - 3/24 < 1
13/24 < 1 (True).
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Answer:
The answer is 10x-26 or try using the Distributive Property.
Step-by-step explanation:
Answer:
Step-by-step explanation:
You have to multiply 30 cents until you get to the total.
There are 28 students in first period and 56 students in sceond period and 74 students in third period
<em><u>Solution:</u></em>
Let the number of students in first period be "x"
Let the number of students in second period be "y"
Let the number of students in third period be "z"
<em><u>There are 158 students registered for American History classes.</u></em>
Therefore,
x + y + z = 158 ---------- eqn 1
<em><u>There are twice as many students registered in second period as first period</u></em>
number of students in second period = twice of number of students in first period
y = 2x ------- eqn 2
<em><u>There are 10 less than three times as many students in third period as in first period</u></em>
number of students in third period = 3 times number of students in first period - 10
z = 3x - 10 ------ eqn 3
<em><u>Substitute eqn 2 and eqn 3 in eqn 1</u></em>
x + 2x + 3x - 10 = 158
6x = 168
<h3>x = 28</h3>
<em><u>Substitute x = 28 in eqn 2</u></em>
y = 2(28)
<h3>y = 56</h3>
<em><u>Substitute x = 28 in eqn 3</u></em>
z = 3(28) - 10
z = 84 - 10
<h3>z = 74</h3>
Thus there are 28 students in first period and 56 students in sceond period and 74 students in third period