1 1 1 2
2 -3 1 -11 -2R1 + R2 → R2
-1 2 -1 8 R1 + R3 → R3
1 1 1 2
0 -5 -1 -15 R2 ⇔ R3
0 3 0 10
1 1 1 2
0 3 0 10 -R3
0 -5 -1 -15
1 1 1 2
0 3 0 10 1/3 R2
0 5 1 15
1 1 1 2 -R2 + R1
0 1 0 10/3 -5R2 + R3
0 5 1 15
1 0 1 -4/3
0 1 0 10/3 -R3 + R1
0 0 1 -5/3
1 0 0 1/3
0 1 0 10/3
0 0 1 -5/3
Therefore, x = 1/3, y = 10/3, z = -5/3
Answer:
Multiply (400 + 20 + 3) x 10000
Add 4 zeros to the end of 423
Step-by-step explanation:
423 x 10,000 = Adding the amount of zero’s in 10000 to 423
Question 21
Let's complete the square
y = 3x^2 + 6x + 5
y-5 = 3x^2 + 6x
y - 5 = 3(x^2 + 2x)
y - 5 = 3(x^2 + 2x + 1 - 1)
y - 5 = 3(x^2+2x+1) - 3
y - 5 = 3(x+1)^2 - 3
y = 3(x+1)^2 - 3 + 5
y = 3(x+1)^2 + 2
Answer: Choice D
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Question 22
Through trial and error you should find that choice D is the answer
Basically you plug in each of the given answer choices and see which results in a true statement.
For instance, with choice A we have
y < -4(x+1)^2 - 3
-7 < -4(0+1)^2 - 3
-7 < -7
which is false, so we eliminate choice A
Choice D is the answer because
y < -4(x+1)^2 - 3
-9 < -4(-2+1)^2 - 3
-9 < -7
which is true since -9 is to the left of -7 on the number line.
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Question 25
Answer: Choice B
Explanation:
The quantity (x-4)^2 is always positive regardless of what you pick for x. This is because we are squaring the (x-4). Squaring a negative leads to a positive. Eg: (-4)^2 = 16
Adding on a positive to (x-4)^2 makes the result even more positive. Therefore (x-4)^2 + 1 > 0 is true for any real number x.
Visually this means all solutions of y > (x-4)^2 + 1 reside in quadrants 1 and 2, which are above the x axis.
Answer:
9
Step-by-step explanation:
Solution
Here,
(a^3)^6=(a^2)^x
or, a^18=a^2x
or, 18=2x
or, x=18/2
x=9
