Answer:
(Choice C) C Replace one equation with a multiple of itself
Step-by-step explanation:
Since system A has the equations
-3x + 12y = 15 and 7x - 10y = -2 and,
system B has the equations
-x + 4y = 5 and 7x - 10 y = -2.
To get system B from system A, we notice that equation -x + 4y = 5 is a multiple of -3x + 12y = 15 ⇒ 3(-x + 4y = 5) = (-3x + 12y = 15).
So, (-x + 4y = 5) = (1/3) × (-3x + 12y = 15)
So, we replace the first equation in system B by 1/3 the first equation in system A to obtain the first equation in system B.
So, choice C is the answer.
We replace one equation with a multiple of itself.
A. turn all fractions into decimals or all decimals into fractions; I've decided to turn all fractions into decimals
525
------- = 0.525
1000
3
-- = 0.75 = 0.750 ~ (add a zero at the end to make the decimal places 4 equal with 0.525)
0.55 = 0.55 = 0.550 ~<span>(add a zero at the end to make the decimal places equal with 0.525 and 0.750)
</span>
<span> Answer to Problem 'a':</span>
Smallest = 0.525 = 525
-------
1000
Middle = 0.550 = 0.55
Largest = 0.750 = 3
--
4
b. <span>turn all fractions/mixed fractions into decimals or all decimals into fractions; I've decided to turn all fractions/mixed fractions into decimals
</span>
3.805 = 3.805
3.85 = 3.850 ~ ( <span>add a zero at the end to make the decimal places equal with 3.805)
</span>
3 4/5 = 80
----- = 0.80 =0.800 ~ (add a zero at the end to make the decimal
100 places equal with 3.805 and 3.850)
Answers to Problem 'b':
Smallest = 0.800 = 0.80 = 3 4/5
Middle = 3.805 = 3.805
Largest = 3.850 = 3.85
64 = 2 ^6
16 = 2^4 =
2^ 6*(2x+4) = 2^4 * 5x
6(2x+4) = 20x
10/3 x = 2x +4
4/3 x = 4
x = 3
you can also use logarithms like so
(2x+4)ln64 = 5x ln16
ln64/ln16 = 3/2
3/2 * 2x + 3/2 * 4 = 5x
3x + 6 = 5x
2x = 6
x = 3
Will start to distribute the 2 in (x-7)
so it can be (2x-2*7)=100
(2x-14)=100
2x=100+14
2x=114
x=114/2
x=57
