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joja [24]
2 years ago
10

Simplify

c{6a^2 b^-^2}{8a^-^3 b^3}" alt="\frac{6a^2 b^-^2}{8a^-^3 b^3}" align="absmiddle" class="latex-formula"> Assume a≠0 and b≠0

Mathematics
2 answers:
romanna [79]2 years ago
4 0

Answer:

3rd option

Step-by-step explanation:

using the rules of exponents

\frac{a^{m} }{a^{n} } = a^{(m-n)} : nm > n

\frac{a^{m} }{a^{n} } = \frac{1}{a^{(n-m)} } : n > m

\frac{6a^2b^{-2} }{8a^{-3b^3} } ← separate the variables

= \frac{6}{8} × \frac{a^2}{a^{-3} } × \frac{b^{-2} }{b^3}

= \frac{3}{4} × a^{2-(-3)} × \frac{1}{b^{3-(-2)} }

= \frac{3}{4} × a^{2+3} × \frac{1}{b^{3+2} }

= \frac{3}{4} × a^{5} × \frac{1}{b^{5} }

= \frac{3a^{5} }{4b^{5} }

vredina [299]2 years ago
4 0

Answer:

sorry i thought i knew it

Step-by-step explanation:

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Hi Genius~<br><br> Can you help me please ? Thanks! Please show and explain your work :)
Murljashka [212]
The translation that maps triangle ABC to A prime B prime C prime would be a reflection across the y axis. This is because when you reflect something, you are pretty much flipping it. When you reflect across the y axis, you are flipping the triangle across the y axis. Take one point for example. I will use C. Notice how the point C is 3 units away from the y axis. So the same way you would move the point 3 units right from the y axis, and that would be your new point. This sounds kind of complicated, so I will give you a list of rules to make it more simple.


Reflection across y axis: (x,y) would be equal to (-x, y)
Reflection across x axis: (x,y) would be equal to (x, -y)
Reflection across y = x: (x,y) would be equal to (y,x)
Reflection across y = x: (x,y) would be equal to (-y,-x).

A reflection across y = x would be when you have a line that for every 1 it rises, it goes right 1. It is a positive line, as opposed to the y = -x line. It also has a slope of 1. I will try attaching a graph if I can.

Anyway, as I was saying. So pretty much if you don't want to go through the logic, to see whether a figure is reflected, just try each of these rules and if one works then you have your answer. Otherwise it would not be a reflection.

Thanks for being a great mod and hope this helps! :D
5 0
3 years ago
An ellipse has a center at the origin, a vertex along the major axis at (0, 17), and a focus at (0, −8).
MAVERICK [17]

Answer:

The last one.

Step-by-step explanation:

The standard form equation of an ellipse with foci on the y-axis is ...

  y²/a² +x²/b² = 1

where "a" and "b" are the lengths of the semi-major and semi-minor axes, respectively. If "c" is the distance from the center to the focus, then this relation also holds:

  b² +c² = a²

For this ellipse ...

  b² + 8² = 17²

  289 -64 = b² = 225 . . . . . subtract 8²

  b = 15 . . . . . . . . . . . . . . . . . take the square root

The equation of the ellipse is then ...

  y²/17² +x²/15² = 1

3 0
3 years ago
Read 2 more answers
2/-7 in standard form​
Elan Coil [88]

Answer:

the answer is

x2 - 7 =0

Step-by-step explanation:

no explination

5 0
3 years ago
–3h − 3h − –15h − 6 = –15<br> Solve for h <br> I WILL GIVE BRAINLIEST &lt;33
salantis [7]

Answer:

Isolate the variable by dividing each side by factors that don't contain the variable.

h= -1

7 0
3 years ago
Evaluate the function f(x) = 3x2− 2x for x= 4.
kykrilka [37]

Answer:

40

Step-by-step explanation:

f(4) = 3(4²) - 2(4)

= 3(16) - 8

= 48 - 8

= 40

when x=4, f(x) = 40

6 0
3 years ago
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