Answer:
2/27
Explanation:
• Number of green marbles = 2
,
• Number of blue marbles = 4
,
• Number of yellow marbles = 3
Total = 2+4+3 =9
• Probability of choosing a green marble = 2/9
,
• Probability of choosing a yellow marble = 3/9
Therefore, the possibility that the first marble is green and the second is yellow is:
To obtain the square root of 16x^36, the coefficient portion (16) will not present any problems since 16 is a perfect square. However, for a variable with an exponent, the exponent is to be multiplied by 1/2 since the square root symbol is equal to raising the term inside to the power of 1/2. This is shown below:
sqrt (16 x^36) = 4 * x^36(1/2) = 4 * x^18
Therefore, the correct answer is 4x^18.
Answer:
The equation for finding the hypotenuse is:
c^2+c^2=h^2
For example, in the first exercise:
8^2+10^2=164^2
For clearing the equation, find out the root of 164=
12.8
Hypotenuse of the first triangle is 12.8!
Let me help you with the next, if you still dont get it:
8^2+13^2=233^2
Root of 233: 15.2
Hypotenuse of second triangle is 15.2!
Answer:
x = 40°
Step-by-step explanation:
Two sides are the same length, so the angles opposite them are the same size.
The sum of the angles = 180°. One angle is 100°, so the sum of the remaining two angles is 80°.
Since the remaining two angles are congruent, each is 40°.
In the first octant, the given plane forms a triangle with vertices corresponding to the plane's intercepts along each axis.



Now that we know the vertices of the surface

, we can parameterize it by

where

and

. The surface element is

With respect to our parameterization, we have

, so the surface integral is