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Phantasy [73]
3 years ago
7

Jefferson school has students in 1st grade up to 5st grade. The number of children in 1st grade has 3 digits.The digit in the nu

mber are 2,3,and 8 .The digit 8 means 80 in the number write the number that could be the number of children in 1st grade.
Mathematics
2 answers:
lapo4ka [179]3 years ago
7 0

Answer:

283 or 382

Step-by-step explanation:

The number of children in 1st grade has 3 digits.The digit in the number are 2,3,and 8 .

Since the digit 8 means 80 in the number. The other two numbers could interchange the Hundred and Unit position respectively.

The possible numbers are:

I. 200+80+3=283

II. 300+80+2=382

The possible number of children in 1st Grade is either 283 or 382.

sladkih [1.3K]3 years ago
7 0

Answer:

283 or 382

Step-by-step explanation:

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Gemiola [76]

Answer:3/14

Step-by-step explanation:

The probability that it is a blue marble is 3 out of total marbles of 14

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3 years ago
Which number is NOT an integer?
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I believe the answer is C
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I NEED HELP WITH THIS HELP PLS!!!!
motikmotik

Answer:

120

Step-by-step explanation:

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4 0
3 years ago
i have been trying to solve this compound and double angle question please help me find the answer to these question guys​
natali 33 [55]

Answer:

These type of questions are super tricky b/c you have to remember all the different versions of the identities, and then they put the question in some odd form,  I feel like this should land math professors in jail , for dishonesty , b/c it's really a form of "how tricky can I make a question and still have a way to solve it"   anyway,

Step-by-step explanation:

a)

next the question asks   1-cos 2A   and this is total abuse of notation.   the way this should be written is   1- cos( 2A)  so we know that the A is part of the cosine functions input... btw.. in any computer program,  it would never ever let you get away with that top form of the expression.  :/   anyway... I keep ranting.. huh... sorry  :P

1-cos(2A) is an odd form of the identity  1/2(1-cos(2A) = sin^{2}(A)  the 1/2 is missing but we can add that pretty easy, we just have to remember to take it out too. I usually forget to do that. and my professor marks me off completely,  totally wrong, but I just miss one small thing  :/  anyway....

our 1-cos(2A) needs the 1/2 added to it.  or if we move that 1/2 to the other side it looks like  2*sin^{2}(A)  = 1-cos(2A)  and this is that "odd" from of the identity that I was talking about.  

next let's deal with sin(2A)  it has an identity of  2 sin(A)cos(A) which is really nice for us b/c it will cancel out the 2 in then numerator for us, nice !

now our fraction looks like  [2* sin^{2}(A)] / 2 sin(A)cos(A)

so cancel out one of the sines

2*sin(A) / 2 cos(A)

cancel the 2s

Sin(A) / Cos(A) = Tan(A)

nice  it worked out  :P

b)

by the above that we just worked out, then

Tan(15) = Sin(15) / Cos(15)

I had to look up what sin of 15 is b/c it's not one of those special angles but it does have an exact form of

Sin(15) = (√3 - 1) / 2√2

Cos(A) = (√3 + 1) / 2√2

you can use rule of Cos(A-B) = Cos(A)Cos(B)+Sin(A)Sin(B) to get the above and a similar rule for Sin(A-B)

back to our problem,  the 2√2 will cancel out

then we have

Tan(15) =  (√3 - 1) /(√3 + 1)

in the form that is above that's exact, the roots could be approximated but i'll just leave that in the form that is exact.  Most math professors like that form.  

 

4 0
2 years ago
4. Is the point (3,-4) a solution to the given system? Justify your answer.
Zigmanuir [339]

Answer: No, it is not a solution

==========================================================

Explanation:

The point (3,-4) means that x = 3 and y = -4 pair up together

Let's plug these x,y values into each equation

Starting with the first equation, we get,

y = 4x-16

-4 = 4(3)-16 ... x replaced with 3; y replaced with -4

-4 = 12-16

-4 = -4 .... this is a true statement

Repeat for the second equation

y = 2x-6

-4 = 2(3)-6

-4 = 6-6

-4 = 0 ... this is false

Since we get a false statement, this means (3,-4) is not on the line y = 2x-6, which means that overall (3,-4) is not a solution to the system of equations. The point (3,-4) must make both equations true for it to be a solution.

5 0
3 years ago
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