Answer:
1. Positive, 1+2=3
2. Negative, -1-2=-3
Step-by-step explanation:
If you look at both in a graphing perspective, the point (1,2) is in Quadrant I. likewise, adding 2 to the x-coordinate will also result in the point (3,2), also in Quadrant I, where the x coordinate is positive. The point (-1,2) is in Quadrant II, and adding -2 to the x coordinate keeps it in Quadrant II, where the x-coordinate is negative.
Answer:
The second option: 3 (6 - 5n)/20n
Step-by-step explanation:
Make sure all fractions have a common denominator:
Step 1. Find a common multiple between all three denominators
5, 4, and 10 all have a common multiple of 20. Proof: 5 × 4 = 20, 4 × 5 = 20, and 10 × 2 = 20
Step 2. Multiply the denominators to get to 20. Whatever you do to the bottom (denominator) must be done to the top (numerator).
1/5n × 4/4 = 4/20n
3/4 × 5n/5n = 15n/20n
7/10n × 2/2 = 14/20n
Your fractions now all have a common denominator of 20n.
Rewrite the equation using the new fractions:
4/20n - 15n/20n + 14/20n
Only focus on adding/subtracting the numerators; the denominators will stay the same: 20n.
(4 - 15n + 14)/20n
Combine like terms:
(18 - 15n)/20n
Factor out any numbers possible:
3(6 - 5n)/20n
Note* 3 go into both 18 and 15, which allows us to factor 3 out. 18 ÷ 3 = 6 and 15 ÷ 3 = 5, giving us our new numbers inside the parentheses.
Answer:
length times width times height
Step-by-step explanation:
width is height by the way
so 20×10×9 gives you 1800
1800 petrol can hold on Meters
Answer:
For this case we have by definition, that an irrational number is a number that can not be expressed as a fraction \ frac {a} {b}, where a and b are integers and b is different from zero.
A) , it is rational
B), it is rational
C) , it is rational
D) it is not rational.
Step-by-step explanation:
dddddd its dddd
BESTI ILYSM
The standard form of a quadratic equation is
, while the vertex form is:
, where (h, k) is the vertex of the parabola.
What we want is to write
as
First, we note that all the three terms have a factor of 3, so we factorize it and write:
.
Second, we notice that
are the terms produced by
, without the 9. So we can write:
, and substituting in
we have:
![\displaystyle{ y=3(x^2-6x-2)=3[(x-3)^2-9-2]=3[(x-3)^2-11]](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%20y%3D3%28x%5E2-6x-2%29%3D3%5B%28x-3%29%5E2-9-2%5D%3D3%5B%28x-3%29%5E2-11%5D)
.
Finally, distributing 3 over the two terms in the brackets we have:
![y=3[x-3]^2-33](https://tex.z-dn.net/?f=y%3D3%5Bx-3%5D%5E2-33)
.
Answer:
