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elixir [45]
2 years ago
11

If ray BD bisects angle CBE, ray BC is transparent to ray BA, angle CBD = (3x + 25) degrees, and angle DBE = (7x - 19) degrees,

find angle ABD

Mathematics
2 answers:
Alex777 [14]2 years ago
7 0
Check the picture below.

Alenkasestr [34]2 years ago
5 0

Answer:  The required measure of angle ABD is 148°.  

Step-by-step explanation:  Given that ray BD bisects angle CBE, ray BC is transparent to ray BA.

Also,

m\angle CBD=(3x+25)^\circ,~~m\angle DBE=(7x-19)^\circ,~~m\angle ABC=90^\circ.

We are to find the measure of angle ABD.

Since ray BD bisects angle CBE, so we have

m\angle CBD=m\angle DBE\\\\\Rightarrow (3x+25)^\circ=(7x-19)^\circ\\\\\Rightarrow 3x+25=7x-19\\\\\Rightarrow 7x-3x=25+19\\\\\Rightarrow 4x=44\\\\\Rightarrow x=\dfrac{44}{4}\\\\\Rightarrow x=11.

So, we get

m\angle CBD=(3\times11+25)^\circ=58^\circ.

Therefore,

m\angle ABD=m\angle ABC+m\angle CBD=90^\circ+58^\circ=148^\circ.

Thus, the required measure of angle ABD is 148°.  

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Strike441 [17]
<h2>Answer with explanation:</h2>

Given : An urn contains 2 red marbles and 3 blue marbles.

Total marbles = 2+3=5

a) The general ways in which the person could get a red marble and a blue marble are :

1)  He draws red marble first and then second marble as blue.

2)  He draws blue marble first marble and then second marble as red.

b)  The number of ways to get one red and one blue marble is given by :-

^2C_1\times^3C_1=2\times3=6                          (i)

c) Number of ways to get 2 marbles from 5 is given by :-

^5C_2=\dfrac{5!}{2!(5-2)!}=\dfrac{5\times4\times3!}{3!\times2}=10     (ii)

Now, The probability the person gets a red and a blue marble will be :-

P(R\ \&B)=\dfrac{6}{10}=0.6       [Divide (i) by (ii)]

Hence, the  probability the person gets a red and a blue marble= 0.6

5 0
2 years ago
The owner of a football team claims that the average attendance at games is over 74,900, and he is therefore justified in moving
Slav-nsk [51]

Answer:

Option c (Upper tailed) is the correct choice.

Step-by-step explanation:

Given that:

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= 74,900

We will have to test:

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or,

   H_0: \mu \leq 74,900

Verses,

⇒ H_1: \mu> \mu_0

or,

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The other given alternatives aren't connected to the given scenario. So the above is the correct one.

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Paco's cell phone carrier charges him $0.20 for each text message he sends or receives, $0.15 per minute for calls, and a $15 mo
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Given:

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To Find:

The number of texts 't' Paco can send/receive in a month.

Answer:

0\leq t

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Step-by-step explanation:

Paco wants to keep is monthly bill below $30.

We see that he has to pay a foxed monthly service fee of $15. This means he is only left with a limit of $30 - $15 = $15 for his monthly calls and texts.

That is, the amount he has to pay for texting and calling has to be less than $15.

For texts, the cell phone carrier charges $0.20 for sending/receiving texts.

For calls, he is charged $0.15 per minute.

Let the number of text messages Paco can send or receive in a month be denoted by 't'.

Let the number of minutes Paco can call in a month be denoted by 'c'.

Then, the total cost of text messages he can send or receive per month would be 0.20t and the total cost of the minutes he spends on calls would be 0.15c. Together, the sum of these has to be less than $15 if his monthly bill has to be kept less than $30 (accounting for the monthly service fee).

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0.20t+0.15c

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So, putting c = 0, the aboce equation can be written as

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That is, Paco has to send and receive less than 75 texts.

So,

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3 0
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brilliants [131]

Answer:

The meet at first time at starting point after 336 seconds,

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Given that Venu and Tanuj take 48 seconds and 56 seconds to

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To meet them again at the startiong point, the  time taken must be multiple of 56 and 48.

So, the minimum time required to meet them again at the startiong point is time taken for first time meet.

Time taken to first time meet is the smallest number which can be divided by 56 and 48, i.e the lowest common factor (LCM) of 56 and 48 which is 336 seconds or 5 minutes 36 seconds.

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Hence, the  change in Grandpa Ernie's height each year = 0.6 cm

4 0
2 years ago
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