1. Write your answer using only positive exponents. (3ab^3c^-1)^3

Mathematics

2 answers:

pishuonlain [190]3 years ago

7 0

Maybe it is 27b^9a^3 over c^3

marishachu [46]3 years ago

6 0

Answer:

$(\frac{{3ab^{3}}}{c^{1}} )^{3}$

Step-by-step explanation:

$(3ab^{3} c^{-1} )^{3} =(\frac{{3ab^{3}}}{c^{1}} )^{3}$

You might be interested in

The arc of the parabola y = x^2 from (3,9) to (4,16) is rotated about the y-axis. Find the area of the resulting surface. Please

Hitman42 [59]

Answer:

156.5

Step-by-step explanation:

Thinking process:

The area can be calculated using the formula:

$S = \int\limits^4_3 {2\pi x\sqrt{1+}(2x)^{2} } \, dx$

We let the substitution take place.

Therefore, we let $u = 1 + 4x^{2}$

Thus, $du = 8dx.$

So,

$xdx = \frac{1}{8}du$

Also, the interval of the integration changes to [ 37, 65]

Thus,

$S = \int\limits^4_3 {2\pi \sqrt{1+4x^{2} } } \, dx \\= \int\limits^65_35 {2\pi \sqrt{\frac{1}{8}u } } \, dx$

= $\frac{1}{6} [ 65^{\frac{3}{2}-37^{\frac{3}{2} } }$

= 156.5 units²

8 0

3 years ago

Triangle J K L is shown. Angle J L K is 105 degrees. The length of J K is 4.7 and the length of J L is 2.7.

cluponka [151]

Answer:

The approximate measure of angle K is 34°

Step-by-step explanation:

In ΔJKL

∠L = 105°

JK = 4.7

JL = 2.7

Sine Rule:

$\frac{SinA}{a} = \frac{SinB}{b}= \frac{SinC}{c}$

So, $\frac{SinL}{JK} = \frac{SinK}{JL}\\\frac{SinL}{4.7} = \frac{SinK}{2.7}\\\frac{Sin 105}{4.7} = \frac{SinK}{2.7}\\\frac{Sin 105 \times 2.7}{4.7} = SinK\\\frac{(0.9659 \times 2.7)}{4.7} = Sin K\\\frac{2.60793}{4.7} = Sin K\\0.5549 = Sin K\\Sin^{-1}(0.5549)= K\\K = 33.70$