When using mental math to solve problems with decimals, it is harder to keep track of what number goes where and where the decimal spot goes as well. Although, with whole numbers in mental math, it is a lot easier to solve since there is no decimal and less numbers (unless you are doing whole number math with a large amount of numbers).
<h3>
Answer: 8.2 (choice B)</h3>
==========================================================
Explanation:
First we'll need the length of segment BD, which I'll call x for now. The similar triangles allow us to set up the proportion below to solve for x
AD/BD = BD/DC
13/x = x/4
13*4 = x^2
52 = x^2
x^2 = 52
x = sqrt(52)
Segment BD is exactly sqrt(52) units long.
------------
Now focus your attention on triangle BDC. We'll use the pythagorean theorem to find BC
(BD)^2 + (DC)^2 = (BC)^2
( sqrt(52) )^2 + ( 4 )^2 = ( m )^2
52 + 16 = m^2
68 = m^2
m^2 = 68
m = sqrt(68)
m = 8.2462 approximately
m = 8.2
Answer:
let me know
Step-by-step explanation:
Answer:
Step-by-step explanation:
To compare the fractions we require them to have a common denominator.
The lowest common multiple of 3 and 7 is 21
Change both fractions into fractions with a denominator of 21
and 
= 
and 
The numerator 14 is greater than 12, hence
is the larger fraction
Let’s take a look at the definitions of all of these categories of polygon:
A rhombus is a quadrilateral with four equal-length sides
A trapezoid is a quadrilateral with at least one pair of parallel sides
A kite is a quadrilateral with two pairs of equal sides, where those sides are adjacent to each other
A quadrilateral is a four-sided polygon
And let’s compare our definitions with the figure:
- None of the sides of the figure are equal to each other, so we it can’t be a rhombus
- The slopes of all of the sides are different, so the figure can’t have a pair of parallel lines, ruling out the chance that it’s a trapezoid
- That first bullet point also rules out the possibility that our figure is a kite
- Our figure *is* a four-sided polygon though, so it meets the requirements for a quadrilateral
So, the only label that works for this figure is a *quadrilateral*.
