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3 years ago

Do ΔCBG and ΔFEG in the figure appear to be similar? Why or why not?

Mathematics

2 answers:

astra-53 [7]3 years ago

8 0

Answer:

Step-by-step explanation:

They're not similar because only one pair of corresponding angles in the two triangles is congruent.

Mama L [17]3 years ago

6 0

Answer: Option C. They're not similar because only one pair of corresponding angles in the two triangles is congruent.

Solution:

The only one pair of corresponding angles in the two triangles is congruent is the angle BGC in ΔCBG and the angle FGE in ΔFEG, because they are vertical angles (opposite by the vertex)

You might be interested in

Quadrilateral PQRS is a square whos side length is 10. Let X and Y be points outside the square so that XQ = YS = 6 and XP = YR

erik [133]

Answer:

392

Step-by-step explanation:

Triangles XQP and YRS are right triangles because triples 6, 8, 10 are Pythagorean triples.

Extend lines XQ, YR, YS and XP and mark their intersection as A and B.

Quadrilateral XAYB is a square because all right triangles PXQ, QAR, RYS and SBP are congruent (by ASA postulate) and therefore

all angles of the quadrilateral XAYB are right angles

all sides of XAYB are congruent and equal to 6 + 8 = 14 units.

Segment XY is the diagonal of the square XAYB, by Pythagorean theorem,

$XY^2=XA^2+AY^2\\ \\XY^2=14^2+14^2\\ \\XY^2=196+196\\ \\XY^2=392$

3 0

2 years ago

Find three consecutive even numbers whose sum is 1674.

Norma-Jean [14]

Answer:

556, 558, and 560

Step-by-step explanation:

1674/3 = 558 (An even number, yay)

So 558 + 558 + 558 = 1674

Subtract two from the first one and add two to the last (still equal)

556 + 558 + 560

adding them equals 1674 so the answer is

556, 558, and 560

7 0

3 years ago

Points (0, 0) and (x, y) ?

kap26 [50]

Answer:

m=y/x

Step-by-step explanation:

8 0

2 years ago

ANSWER THIS AND ILL LOVE YOU <br><br> find angles FAB and BAC

jeyben [28]

Answer:

m<FAB = 75°

m<BAC = 105°

Step-by-step explanation:

First, find the value of x.

(13x - 3)° = (3x + 2)° + 55° (exterior angle theorem of a ∆)

Solve for x

13x - 3 = 3x + 2 + 55

13x - 3 = 3x + 57

Collect like terms

13x - 3x = 57 + 3

10x = 60

Divide both sides by 10

x = 6

✔️m<FAB = 13x - 3

Plug in the value of x

m<FAB = 13(6) - 3 = 78 - 3

m<FAB = 75°

✔️m<BAC = 180 - m<FAB (angles on a straight line/supplementary angles)

m<BAC = 180 - 75 (substitution)

m<BAC = 105°

4 0

2 years ago

In a study, 36% of adults questioned reported that their health was excellent. A researcher wishes to study the health of peop

steposvetlana [31]

Answer:

0.3907

Step-by-step explanation:

We are given that 36% of adults questioned reported that their health was excellent.

Probability of good health = 0.36

Among 11 adults randomly selected from this area, only 3 reported that their health was excellent.

Now we are supposed to find the probability that when 11 adults are randomly selected, 3 or fewer are in excellent health.

i.e. $P(x\leq 3)=P(x=1)+{P(x=2)+P(x=3)$

Formula : $P(x=r)=^nC_r p^r q ^ {n-r}$

p is the probability of success i.e. p = 0.36

q = probability of failure = 1- 0.36 = 0.64

n = 11

So, $P(x\leq 3)=P(x=1)+{P(x=2)+P(x=3)$

$P(x\leq 3)=^{11}C_1 (0.36)^1 (0.64)^{11-1}+^{11}C_2 (0.36)^2 (0.64)^{11-2}+^{11}C_3 (0.36)^3 (0.64)^{11-3}$

$P(x\leq 3)=\frac{11!}{1!(11-1)!} (0.36)^1 (0.64)^{11-1}+\frac{11!}{2!(11-2)!} (0.36)^2 (0.64)^{11-2}+\frac{11!}{3!(11-3)!} (0.36)^3 (0.64)^{11-3}$

$P(x\leq 3)=0.390748$

Hence the probability that when 11 adults are randomly selected, 3 or fewer are in excellent health is 0.3907

5 0

2 years ago