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SIZIF [17.4K]
3 years ago
9

Someone please help me answer this geometry question!! will give brainliest and 5 stars

Mathematics
1 answer:
FromTheMoon [43]3 years ago
3 0
Do 6/24 = x/20 and then cross multiply :)
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Which function has a domain of all real numbers
Pachacha [2.7K]

Answer:

Most of the functions we have studied in Algebra I are defined for all real numbers. This domain is denoted . For example, the domain of f (x) = 2x + 5 is , because f (x) is defined for all real numbers x; that is, we can find f (x) for all real numbers x.

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
What is the answer? I don’t know how to say that much nicer.. i need help though.
Marysya12 [62]
Sorry bro idkhdjdjjdjdjdjfjfjxjj83848 89**%%
5 0
3 years ago
Help please ASAP!!!! I NEED CORRECT ANSWERS ONLY FOR BRAINLIEST
MrRissso [65]

Let us start subtracting 3 cos theta both sides so that we may get all theta terms on left side only

It is now : 4 cos theta - 3 cos theta = - (sqrt3)/ 2

1 cos theta = -( sqrt 3 ) /2

we know cos (pi-x)= - cos x And cos (pi+ x)= - cos (x)

********* let us first use :

- Cos ( pi - theta) = - ( sqrt 3 )/2

this implies cos (pi- theta )= (sqrt 3) /2

cos ( pi - theta )= cos (pi /6)

pi- theta = pi/6

pi- pi/6 = theta

5 pi/6 = theta

one of teh answer is 5 pi/6 . another value of theta would be :

(2pi - 5pi/6)= 7pi/6

Answer: 5 pi/6 and 7 pi/6

5 0
3 years ago
The sum of mZAOB and mZBOC IS 128º. Find the measure of ZBOC.
amm1812
128 - 33 = 95
Answer C. 95
4 0
3 years ago
Read 2 more answers
A coin is tossed 5 times. Find the probability that all are heads. Find the probability that at most 2 are heads.
fenix001 [56]

Answer:

1/32

15/32

Step-by-step explanation:

For a fair sided coin,

Probability of heads, P(H) = 1/2

Probability of tails P(T)  = 1/2

For a coin tossed 5 times,

P( All heads)

= P(HHHHH),

= P (H) x P(H) x P(H) x P(H) x P(H)

= (1/2) x (1/2) x (1/2) x (1/2) x (1/2)

= 1/32 (Ans)

For part B, it is easier to just list the possible outcomes for

"at most 2 heads" aka "could be 1 head" or "could be 2 heads"

"One Head" Outcomes:

P(HTTTT), P(THTTT) P(TTHTT), P(TTTHT), P(TTTTH)

"2 Heads" Outcomes:

P(HHTTT), P(HTHTT), P(HTTHT), P(HTTTH), P(THHTT), P(THTHT), P(THTTH), P(TTHHT), P(TTHTH), P(TTTHH)

If we count all the possible outcomes, we get 15 possible outcomes representing "at most 2 heads)

we know that each outcome has a probability of 1/32

hence 15 outcomes for "at most 2 heads" have a probability of

(1/32) x 15  = 15/32

8 0
3 years ago
Read 2 more answers
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