Perimeter of Parallelogram ABCD: Pp=10 cm
BD=?
Perimeter of triangle ABD: Pt=8 cm
Parallelogram ABCD:
AB=CD=L1
BC=AD=L2
Pp=2(L1+L2)
10 cm=2(L1+L2)
Dividing by 2 both sides of the equation:
(10 cm)/2=2(L1+L2)/2
5 cm=L1+L2
L1+L2=5 cm
Triangle ABD
Pt=AB+BD+AD
8 cm=L1+BD+L2
8 cm=BD+L1+L2
But L1+L2=5 cm, then:
8 cm=BD+5 cm
Solving for BD:
8 cm-5 cm=BD+5 cm+ 5 cm
3 cm=BD
BD=3 cm
Answer: The length of the diagonal BD is 3 cm
Answer:
18
Step-by-step explanation:
![\bf \begin{array}{lccclllll} &amount(g)&\textit{\% of titanium}&\textit{grams of titanium}\\ &-----&-----&-------\\ \textit{6\% alloy}&x&0.06&0.06\cdot x\\ \textit{20\% alloy}&y&0.2&0.2\cdot y\\ ----&-----&-----&-----\\ mixture&100&0.13&100\cdot 0.13\to 13 \end{array}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Blccclllll%7D%0A%26amount%28g%29%26%5Ctextit%7B%5C%25%20of%20titanium%7D%26%5Ctextit%7Bgrams%20of%20titanium%7D%5C%5C%0A%26-----%26-----%26-------%5C%5C%0A%5Ctextit%7B6%5C%25%20alloy%7D%26x%260.06%260.06%5Ccdot%20%20x%5C%5C%0A%5Ctextit%7B20%5C%25%20alloy%7D%26y%260.2%260.2%5Ccdot%20%20y%5C%5C%0A----%26-----%26-----%26-----%5C%5C%0Amixture%26100%260.13%26100%5Ccdot%200.13%5Cto%2013%0A%5Cend%7Barray%7D)
so.. .whatever those amounts are, of "x" and "y", for the mixture,
they must add up to 100 grams, or x + y = 100
and the titanium percentage of that sum, must yield a 13 grams of titanium on those 100 grams
thus
![\bf \begin{cases} x+y=100\to y=\boxed{100-x} \\\\ 0.06x+0.2\boxed{y}=13 \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Ax%2By%3D100%5Cto%20y%3D%5Cboxed%7B100-x%7D%0A%5C%5C%5C%5C%0A0.06x%2B0.2%5Cboxed%7By%7D%3D13%0A%5Cend%7Bcases%7D)
do the substitution, and solve for "x",
what's "y" amount? well, y = 100 - x
60 x 20/100 = 12..........................
A line segment would be C.
This is because it is quite lterally a "segment" of line. Or a piece.