This is the solution:
(3m^-2 n)^-3 / 6mn^-2
For the first step: apply the power distribution3^-3 m^-2*-3 n^-3 / 6mn^-2
In the second step: utilize the product and quotient rulesm^6 n^2/ 3^3 *6*m*n^3
Lastly, simplify the expressionm^5/162n
The conclusive answer is m^5/162n
Hope this aids you.:)
x y
1 290
2 280
3 270
4 260
5 250
6 240
7 230
8 220
9 210
10 200
11 90
12 180
13 170
14 160
15 150
16 140
17 130
18 120
19 110
20 100
21 90
22 80
23 70
24 60
25 50
26 40
27 30
28 20
29 10
30 0
To find the maximum number of identical packs we see we have 72 pencils and 24 calculators.
This involves discovering the largest number that divides both 72 and 24 evenly,
which is known as the GCM or greatest common multiplier.
To determine the GCM, factor 72 into primes and group them:
72=2 times 2 times 2 times 3 times 3
24=2 times 2 times 2 times 3
Thus, the common grouping is 2 times 2 times 2 times 3, equating to 24.
Therefore, the maximum number of packs is 24.
For pencils:
72 divided by 24=3
Resulting in 3 pencils per pack.
For calculators:
24 divided by 24=1
So, 1 calculator per pack.
The outcome is 3 pencils and 1 calculator in each pack.
Answer:
The linear equations are:
Step-by-step breakdown:
Here, a denotes the quantity of 2-point shots, and b indicates the number of 3-point shots.
According to the information given:
The basketball team accumulated 82 points through 2-point and 3-point baskets.
⇒
It is also stated that: They made a total of 38 baskets.
⇒
therefore, the linear equations are:
(a) 4 <span>(b) y = sqrt(9 - (9/16)x^2) </span>The most accurate assumption for the equation based on the general format for an ellipse is: [[TAG_3]]x^2/16 + y^2/9 = 1 [[TAG_4]](a). An ellipse is symmetrical along both its major and minor axes. Thus, if you can calculate the area of the ellipse in one quadrant, multiplying that area by 4 will yield the total area of the ellipse, confirming the factor of 4 is accurate. [[TAG_5]] (b). The standard equation for an ellipse doesn't adequately represent a general function as it results in two y values for each x value. However, if we constrain ourselves to the positive square root, that issue can be resolved easily. Here’s how: [[TAG_6]] x^2/16 + y^2/9 = 1 [[TAG_7]] x^2/16 + y^2/9 - 1 = 0 [[TAG_8]] x^2/16 - 1 = - y^2/9 [[TAG_9]] -(9/16)x^2 + 9 = y^2 [[TAG_10]] 9 - (9/16)x^2 = y^2 [[TAG_11]] sqrt(9 - (9/16)x^2) = y [[TAG_12]] y = sqrt(9 - (9/16)x^2)
