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Vaselesa [24]
3 years ago
13

Problem of the Day:

Mathematics
1 answer:
Shtirlitz [24]3 years ago
8 0

Answer:

Step-by-step explanation:

What this question is asking of you is what is the greatest common divisor of 12 and 15. Or, what is the biggest number that divides both 12 and 15.

in order to find this we have to split each number into it's prime components.

for 12 they are 2,2 and 3 (

2

⋅

2

⋅

3

=

12

)

and for 15 they are 3 and 5 (

3

⋅

5

=

15

)

Out of those two groups (2,2,3) and (3,5) the only thing in common is 3, so 3 is the greatest common divisor. That tells us that the greatest number of groups that can exist and have the same number of girls and the same number of boys for each group is 3.

Now to find out how many girls and boys there are going to be in each group we divide the totals by 3, so:

12

3

=

4

girls per group, and

15

3

=

5

boys per group.

(just as a thought exercise, if there were 16 boys, the divisors would have been (2,2,3) and (2,2,2,2), leaving us with 4 groups [

2

⋅

2

] of 3 girls [12/4] and 4 boys [16/4] )

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A coin is flipped 3 times what is the probability that heads appears twice
blondinia [14]
Note that the chances of a head and a tail appear in a single toss is 1/2
(1/2)^3 =0.125
7 0
3 years ago
Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process!
IgorC [24]
Hey there, hope I can help!

-4x^2+9y^2+32x+36y-64=0

\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \  9y^2+32x+36y-4x^2=64

\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \  -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4
-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9
-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}

\mathrm{Convert}\:x\:\mathrm{to\:square\:form}
-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)

\mathrm{Convert\:to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)

\mathrm{Convert}\:y\:\mathrm{to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)

\mathrm{Convert\:to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)

\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \  -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1

Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1

For me I used
\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1
As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1

Hope this helps!
4 0
3 years ago
What is the solution of the system of equations? y= -3x+8<br> y= -5x-2
Crazy boy [7]
Y = -3x+8
y = -5x-2
------------
y = 5(-3x+8)
y = 3(-5x-2)
---------------
y = -15x+40
y = -15x-6
--------------
y = -15x+40       
       -      - 
y = -15x-6
--------------
y = 46
<span>
y = -3x+8
</span>46 = -3x+8
-8           -8
---------------
38 = -3x
----    ----
-3      -3
12.666...=x
(<span>12.666.,46) </span>← Answer
5 0
3 years ago
Read 2 more answers
V)<br> a common multiple of 6 and 8,<br> What number is a common multiple of 6 and 8?
Vaselesa [24]

Answer:

6, 12, 18, 24, 30, 36, etc.

4 0
3 years ago
Read 2 more answers
Simplify the expressions:
mestny [16]

Answer:

-1v

0.6y+2.1

3w-p+4

1-6x

0

-5x+10y+z-25

Step-by-step explanation:

Combine like terms

8 0
3 years ago
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