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3 years ago

Geometry .........i need ur help lol

Mathematics

2 answers:

poizon [28]3 years ago

8 0

Okay so for 20 you can pretty much do freestyle
I can see arc AD, arc AB, arc CB
Can you name three more??
BTW do you know what an arc looks like? Its almost like a half circle.
So how many half circles can you spot in the picture. I gave you three, can you list three more?

Aloiza [94]3 years ago

6 0

First one is A
Second one is also A
the third one is also A!!!!
DA, AB, BC, CD
square, are side equal length and 90 degree angles
rectangle, two pairs of same length and 90 degree angles
parallelogram, two pairs of same length and two pairs of same angles
rhombus, all sides equal, two pairs of same angles.
yes, because they have the same length of all sides, same shape and same angles.
hope that helps! :D

You might be interested in

Find f(-2) for f(x)=3×2x

Ronch [10]

substitute every x in the equation with (-2)

f(-2)=3*2*(-2) = -12

dunno if I get your equation correctly, pls check how you have written it

7 0

3 years ago

Which represents the solution of the graphed system of equations, y=x^2-2x and y=-2x-1

elena55 [62]

Answer:

x = √(-1) = i

Since the value of √(-1) is not real.

The system has no real solution.

Step-by-step explanation:

The solution to the system of equations is at the point where they intercept each other.

y1 = y2

For the given equation;

y=x^2-2x and y=-2x-1

To get the where they intercept, we will equal both equations;

y=x^2-2x = -2x-1

x^2 - 2x = -2x - 1

x^2 - 2x + 2x + 1 =0

x^2 +1 = 0

x^2 = -1

x = √(-1) = i

Since the value of √(-1) is not real.

The system has no real solution.

6 0

4 years ago

Based on probability and if it is appropriate for a decision to be left up to chance, choose every situation that is both statis

svet-max [94.6K]

The situations that are both statistically fair and morally fair are Situation 2, Situation 3 and Situation 5

<h3>How to categorize the situations?</h3>

For a situation to be fair, the probability of every outcome must be equal.

We can now analyze each option using the above highlight

Situation 1

Alphabets = 26
Probability of vowel = 5/26
Probability of consonant = 21/26

The probabilities of vowels and consonant are not equal.

Hence, this situation is not fair

Situation 2

People = 4

P(Each) = 1/4

The probability of each participant is equal in the above scenario i.e. 1/4

Hence, this situation is fair

Situation 3

People = 5
P(Each) = 1/5

The probability of each cousin is equal in the above scenario i.e. 1/5

Hence, this situation is fair

Situation 4

Numbers = 5 i.e. 2 to 6
P(Prime) = 3/5
P(Composite) = 4/6

The probabilities of prime and composite numbers are not equal.

Hence, this situation is not fair

Situation 5

Roommates = 5
Cards = 5
Aces = 4
P(Each) = 1/5

All probabilities are equal in the above scenario i.e. 1/5

Hence, this situation is fair