I think #13 is A (90 cubic meters) even though you already have that
Okay lets start with "what is a kite?" A kite is a shape which has 4 sides of which 2 pairs of two sides are equal and the center angle is 90 degrees. So mark up your drawing with tick marks to show that segment DF and segment FG are equal (just put a little tick mark on those segments). Similarly segments DH and GH are equal so put two tick marks on them to show they are equal. Mark the center angle with a little square to show it is 90 degrees. Let me give you some hints on the answers. a. is given like you have b. is what you marked on your drawing and is the definition of a kite e. is definition of congruency or even substitution f. Reflexive property - remember it just says that something is congruent to itself. g. looks like the HL property h. you can say that because of CPCTC.
Answer:
![\frac{PQ}{P'Q'}=2.5](https://tex.z-dn.net/?f=%5Cfrac%7BPQ%7D%7BP%27Q%27%7D%3D2.5)
Step-by-step explanation:
we know that
If two figures are similar, then the ratio of its corresponding sides is equal
In this problem Triangles OQP and OQ'P' are similar by AA Similarity Postulate
so
![\frac{PQ}{P'Q'}=\frac{OP}{OP'}](https://tex.z-dn.net/?f=%5Cfrac%7BPQ%7D%7BP%27Q%27%7D%3D%5Cfrac%7BOP%7D%7BOP%27%7D)
we have
![OP=12-2=10\ units](https://tex.z-dn.net/?f=OP%3D12-2%3D10%5C%20units)
![OP'=6-2=4\ units](https://tex.z-dn.net/?f=OP%27%3D6-2%3D4%5C%20units)
substitute
![\frac{PQ}{P'Q'}=\frac{10}{4}=2.5](https://tex.z-dn.net/?f=%5Cfrac%7BPQ%7D%7BP%27Q%27%7D%3D%5Cfrac%7B10%7D%7B4%7D%3D2.5)
Answer:
x=−5+√29 or x=−5−√29
Step
Let's solve your equation step-by-step.
x2+10x+10=14
Step 1: Subtract 14 from both sides.
x2+10x+10−14=14−14
x2+10x−4=0
For this equation: a=1, b=10, c=-4
1x2+10x+−4=0
Step 2: Use quadratic formula with a=1, b=10, c=-4.
x=
−b±√b2−4ac
2a
x=
−(10)±√(10)2−4(1)(−4)
2(1)
x=
−10±√116
2
x=−5+√29 or x=−5−√29
Answer: y=7, x=2
Step-by-step explanation:
∠Q+∠P=180° (∠s between 2 // lines)
97°+(12y-1)°=180°
12y-1=83°
12y=84
y=7
5x+6=8x (Iso triangle)
x=2