<h2>true</h2>
I did it on my calculater and the answer came as 13
hope that helps :)
Answer:
<em>option</em><em> </em><em>A </em><em>is </em><em>correct</em><em> </em><em>!</em>
<em>hope </em><em>this </em><em>answer </em><em>helps </em><em>you </em><em>dear.</em><em>.</em><em>.</em><em>take </em><em>care!</em>
34.
(a) A. x < 4
(b) p > 1
(c) y <span>≤ 2.8
I love solving inequalities. I hope you find this answer the most helpful! :)</span>
First, let's establish a ratio between these two values. We'll use that as a starting point. I personally find it easiest to work with ratios as fractions, so we'll set that up:
To find the distance <em>per year</em>, we'll need to find the <em>unit rate</em> of this ratio in terms of years. The word <em>unit</em> refers to the number 1 (coming from the Latin root <em>uni-</em> ); a <em>unit rate</em> involves bringing the number we're interested in down to 1 while preserving the ratio. Since we're looking for the distance the fault line moves every one year, we'll have to bring that 175 down to one, which we can do by dividing it by 175. To preserve our ratio, we also have to divide the top by 175:
We have our answer: approximately
0.14 cm or
1.4 mm per year
Answer:
0.833 feets
Step-by-step explanation:
Given that:
Volume of box = 1600
Heigh of box (H) = 8 inches
Let:
Width of box (W) = x
Length of box (L) = 2x
Volume of box = (Height * width * length)
1600 = (8 * x * 2x)
1600 = 16x²
x² = 1600/16
x² = 100
x = sqrt(100)
x = 10 inches
Hence, width of box in feet:
1 inch = 0.0833 feets
10 inches = (10 * 0.0833) feets
= 0.833 feets