Please help asap. Question: Triangle FGH is the image of isosceles triangle FEH after a reflection across line HF. Select all the statements that are a result of corresponding parts of congruent triangles being congruent

Diagonal FH bisect angles EFG and EHG and the angle FEH is congruent to angle FGH.

The diagram is given the question.

We need to check all the statements that are applicable for the diagram.

1) EFGH is a rectangle.

The above statements is not true in general. EFGH is a quadrilateral with opposites sides are equal, it may be rectangle in some cases but it is not always be a rectangle.

2). EFGH has 4 congruent sides.

The above statement is also false. EFGH has two opposite sides are congruent.

3). Diagonal FH bisect angles EFG and EHG.

The above statement is true. Triangle EFH and FGH both are isosceles triangles and hence they both have equal angles for the equal sides. Thus, it is true in all cases.

4). Diagonal FH is perpendicular to side Fe.

The above case is not true in general.

5). Angle FEH is congruent to angle FGH.

Yes, the above statement is the correct one, because both the triangles are congruent to each other.

To know more about it, please refer to the link:

link

Options (2), (3) and (5)

Step-by-step explanation:

ΔFGH is the image of isosceles triangle ΔFEH.

Option 1.

EFGH is a rectangle.

False.

(Since measures of all angles are not known)

Option (2).

EFGH has 4 congruent sides.

True.

Option (3).

Diagonal FH bisects angle EFG and EHG.

True.

(Since, ΔFGH is the image of ΔFEH, therefore, both the parts of EFGH will be equal)

Option (4).

Diagonal FH is perpendicular to side FE.

False.

(Since measure of all angles are not known).

Option (5).

∠FEH ≅ ∠FGH

True.

(Since, ΔFEH and  ΔFGH are the image of each other. Therefore, both the triangles will be congruent)

the answer   is 200

step-by-step explanation:

this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12