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3 years ago

Part A

Mathematics

1 answer:

AnnZ [28]3 years ago

7 0

Answer:

blue green and yellow

Step-by-step explanation:

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5) Choose which postulate proves the triangles congruent.

LuckyWell [14K]

Answer:

AAS

Step-by-step explanation:

HEHE

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3 years ago

What is (1/4 -3x^3)^4 in expanded form

Maksim231197 [3]

Answer:

$81x {}^{12} - 27 {}^{9} + \frac{27x {}^{6} }{8} - \frac{3x}{16} + \frac{1}{?256}$

8 0

3 years ago

One urn contains one blue ball (labeled B1) and three red balls (labeled R1, R2, and R3). A second urn contains two red balls (R

marusya05 [52]

Answer:

(a) See attachment for tree diagram

(b) 24 possible outcomes

Step-by-step explanation:

Given

$Urn\ 1 = \{B_1, R_1, R_2, R_3\}$

$Urn\ 2 = \{R_4, R_5, B_2, B_3\}$

Solving (a): A possibility tree

If urn 1 is selected, the following selection exists:

$B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]$

If urn 2 is selected, the following selection exists:

$B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]$

See attachment for possibility tree

Solving (b): The total number of outcome

For urn 1

There are 4 balls in urn 1

$n = \{B_1,R_1,R_2,R_3\}$

Each of the balls has 3 subsets. i.e.

$B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]$

So, the selection is:

$Urn\ 1 = 4 * 3$

$Urn\ 1 = 12$

For urn 2

There are 4 balls in urn 2

$n = \{B_2,B_3,R_4,R_5\}$

Each of the balls has 3 subsets. i.e.

$B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]$

So, the selection is:

$Urn\ 2 = 4 * 3$

$Urn\ 2 = 12$

Total number of outcomes is:

$Total = Urn\ 1 + Urn\ 2$

$Total = 12 + 12$

$Total = 24$

5 0

3 years ago

Solve this equation 6. 14=t -44

Marizza181 [45]

Are you solving for t?

5 0

3 years ago

1) Given: circle k(O), ED= diameter ,m∠OEF=32°, m(arc)EF=(2x+10)° Find: x

qaws [65]

1. The major arc ED has measure 180 degrees since ED is a diameter of the circle. The measure of arc EF is $(2x+10)^\circ$ , so the measure of arc DF is

$m\widehat{DF}=360^\circ-180^\circ-(2x+10)^\circ=(170-2x)^\circ$

The inscribed angle theorem tells us that the central angle subtended by arc DF, $\angle DOF$ , has a measure of twice the measure of the inscribed angle DEF (which is the same angle OEF) so

$m\angle DOF=2m\angle OEF=64^\circ$

so the measure of arc DF is also 64 degrees. So we have

$170-2x=64\implies106=2x\implies\boxed{x=53}$

###

2. Arc FE and angle EOF have the same measure, 56 degrees. By the inscribed angle theorem,

$m\angle EOF=2m\angle EDF\implies56^\circ=2m\angle EDF\implies m\angle EDF=28^\circ$

Triangle DEF is isosceles because FD and ED have the same length, so angles EFD and DEF are congruent. Also, the sum of the interior angles of any triangle is 180 degrees. It follows that

$m\angle EFD+m\angle EDF+m\angle DEF=180^\circ\implies\boxed{m\angle EFD=76^\circ}$

Triangle OFE is also isosceles, so angles EFO and FEO are congruent. So we have

$m\angle EFO+m\angle FEO+m\angle EOF=180^\circ\implies\boxed{m\angle EFO=62^\circ}$

7 0

3 years ago