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vagabundo [1.1K]
3 years ago
5

Find the quotient: (24xy^3-16x^2y^2+32x^2y)/8xy

Mathematics
1 answer:
kolbaska11 [484]3 years ago
5 0
\cfrac{24xy^3-16x^2y^2+32x^2y}{8xy} = \\ \\ \\ = \cfrac{8xy(3y^2-2xy+4x)}{8xy} = \\ \\ \\ 3y^2-2xy+4x
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Anyone know this answer??
lara31 [8.8K]

Answer:

two

Step-by-step explanation:

The exact value of sec(60°) sec ( 60 ° ) is 2 .

6 0
3 years ago
ASAP First to get it right brainliest
nlexa [21]
324

Because you multiply 9 by 36
7 0
2 years ago
Read 2 more answers
How to do substitution or elimination
Levart [38]
I: y=(1/2)x+5
II: y=(-3/2)x-7

substitution:
fancy word for insert the definition of one variable in one equation into the other
-> isolate a variable, luckily y is isolated (even in both equations) already
-> substitute y of II into I (=copy right side of II and replace y in I with it):

(-3/2)x-7=(1/2)x+5
-3x-14=x+10
-3x-24=x
-24=4x
-6=x

-> insert x back into I (or II):
y=(1/2)x+5
=(1/2)*(-6)+5
=-3+5=2

elimination: subtract one equation from the other to eliminate a variable, again y is already isolated->no extra work required

I-II:
y-y=(1/2)x+5-[(-3/2)x-7]
0=(1/2)x+5+(3/2)x+7
0=(4/2)x+12
-12=2x
-6=x

-> insert x back into I (or II):
y=(1/2)x+5
=(1/2)*(-6)+5
=-3+5=2
8 0
3 years ago
What is the solutions to the equation?
victus00 [196]
Hi,

Solving:

x^{3} + 3x^{2} - 4x - 12 = 0 \\ {x}^{2} \times (x + 3) - 4x - 12 = 0 \\ {x}^{2} \times (x + 3) - 4(x - 3) = 0 \\ (x + 3) \times ( {x}^{2} - 4) = 0 \\ x + 3 = 0 \\ {x}^{2} - 4 = 0 \\ \\ (results) \\ x1 = - 3 \\ x2 = - 2 \\ x3 = 2

Hope this helps.
r3t40
4 0
3 years ago
See picture to answer question
SCORPION-xisa [38]

By Cross multiplying given equation will become.

3x-15= 2(4x)

3x -15= 8x

subtracting 3x both sides

3x-3x -15 = 8x-3x

-15 = 5x

Dividing 5 both sides

<u>-5= x   answer</u>

6 0
3 years ago
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