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3 years ago

Read the proof Given: AE EC; BD DC Prove: triangle AEC ~ triangle BDC

Mathematics

2 answers:

Sidana [21]3 years ago

7 0

Answer:

EAC LINE TO~triangle BDC

Step-by-step explanation:

B AND D IS NOT UQAUL

AEC IS THE FIRST GIVEN

SO YOU FINE THE ALL RIGHT ANGLES RIGHT

love history [14]3 years ago

3 0

Answer:

The Answer is A. I just took the test.

Step-by-step explanation:

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the total weight of the medals won by the top medal-winning nations in the 2012 London Olympic Games.

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A survey of 500 farmers showed that of the farmers, 121 grew only wheat, 108 grew only corn, 92 grew only oats, 195 grew wh

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3 years ago

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3 years ago

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Find the value of x in this equation. 25(4x − 3) − 2x = 45 − x

Charra [1.4K]

Answer:

x=40/33

Step-by-step explanation:

Let's solve your equation step-by-step.

25(4x−3)−2x=45−x

Step 1: Simplify both sides of the equation.

25(4x−3)−2x=45−x

(25)(4x)+(25)(−3)+−2x=45+−x(Distribute)

100x+−75+−2x=45+−x

(100x+−2x)+(−75)=−x+45(Combine Like Terms)

98x+−75=−x+45

98x−75=−x+45

Step 2: Add x to both sides.

98x−75+x=−x+45+x

99x−75=45

Step 3: Add 75 to both sides.

99x−75+75=45+75

99x=120

Step 4: Divide both sides by 99.

99x

120

Answer:

x=40/33

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3 years ago

Read 2 more answers

I need answer Immediately pls!!!!!!

Illusion [34]

Given:

Total number of students = 27

Students who play basketball = 7

Student who play baseball = 18

Students who play neither sports = 7

To find:

The probability the student chosen at randomly from the class plays both basketball and base ball.

Solution:

Let the following events,

A : Student plays basketball

B : Student plays baseball

U : Union set or all students.

Then according to given information,

$n(U)=27$

$n(A)=7$

$n(B)=18$

$n(A'\cap B')=7$

We know that,

$n(A\cup B)=n(U)-n(A'\cap B')$

$n(A\cup B)=27-7$

$n(A\cup B)=20$

Now,

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$20=7+18-n(A\cap B)$

$n(A\cap B)=7+18-20$

$n(A\cap B)=25-20$

$n(A\cap B)=5$

It means, the number of students who play both sports is 5.

The probability the student chosen at randomly from the class plays both basketball and base ball is

$\text{Probability}=\dfrac{\text{Number of students who play both sports}}{\text{Total number of students}}$

$\text{Probability}=\dfrac{5}{27}$

Therefore, the required probability is $\dfrac{5}{27}$ .

3 0

3 years ago