2x+1>x-(1-x)
2x+1>x-1+x
2x+1>2x-1
1>-1
Any value of x will make this true.
Answer:
(0 , 0) , (18 , 0) , (18 , -18) , (0 , -18) ⇒ the second answer
Step-by-step explanation:
∵ The vertices of the rectangles are:
(0 , 0) , (0 , 6) , (6 , 6) , (6 , 0)
∵ 3 × ![\left[\begin{array}{ccc}0&1\\-1&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%261%5C%5C-1%260%5Cend%7Barray%7D%5Cright%5D)
× R
∴ That is mean The rectangle rotate 270° around the origin
(270° anti-clockwise or 90° clockwise)
and enlargement by scale factor 3
Answer:
3.4 yards cubed
Step-by-step explanation:
Volume is length x width x height
Length 2 4/5, Width 2, Height 3/5 so 2 4/5 x 2 x 3/5
It looks like they want the answer in decimal form, so you can convert the fractions into decimals.
For the length 2 4/5, you divide the 4 by 5 to get a decimal of .8 so now it's 2.8
For the height of 3/5, you divide the 3/5 & get .6
Now it's 2.8 x 2 x .6 = 3.36 yards cubed which could get rounded up to 3.4 yards cubed
I am a number greater than 40,000 and less than 60,000:
40,000 < n < 60,000
This means that:
n = 10,000n₁ + 1,000n₂ + 100n₃ + 11n₄
And also:
4 ≤ n₁ < 6
0 ≤ n₂ ≤ 9
0 ≤ n₃ ≤ 9
0 ≤ n₄ ≤ 9
My ten thousands digit is 1 less than 3 times the sum of my ones digit and tens digit:
n₁ = 3*2n₄ - 1
n₁ = 6n₄ - 1
This means that:
n = 10,000*(6n₄-1) + 1,000n₂ + 100n₃ + 11n₄
n = 60,000n₄ - 10,000 + 1,000n₂ + 100n₃ + 11n₄
n = 60,011n₄ - 10,000 + 1,000n₂ + 100n₃
<span>My thousands digit is half my hundreds digit, and the sum of those two digits is 9:
n</span>₂ = 1/2 * n₃
<span>
n</span>₂ + n₃ = 9
<span>
Therefore:
n</span>₂ = 9 - n₃
<span>
Therefore:
9 - n</span>₃ = 1/2 * n₃
<span>
9 = 1/2 * n</span>₃ + n₃
<span>
9 = 1.5 * n</span>₃
<span>
Therefore:
n</span>₃ = 6
<span>
If n</span>₃=6, n₂=3.
<span>
This means that:
</span>n = 60,011n₄ - 10,000 + 1,000*3 + 100*6
n = 60,011n₄ - 10,000 + 3,000 + 600
n = 60,011n₄ - 6,400
Therefore:
0<n₄<2, so n₄=1.
If n₄=1:
n = 60,011 - 6,400
n = 53,611
Answer:
53,611
