Answer:
2 < or equal to (t) < or equal to 1000
Step-by-step explanation:
2 is the profit of the (t) amount of t shirts so the amount should be greater than or equal too 1000 because if they have 500 shirts 500 x 2 is 1000
Answer:
3u - 2v + w = 69i + 19j.
8u - 6v = 184i + 60j.
7v - 4w = -128i + 62j.
u - 5w = -9i + 37j.
Step-by-step explanation:
Note that there are multiple ways to denote a vector. For example, vector u can be written either in bold typeface "u" or with an arrow above it
. This explanation uses both representations.
.
.
.
There are two components in each of the three vectors. For example, in vector u, the first component is 11 and the second is 12. When multiplying a vector with a constant, multiply each component by the constant. For example,
.
So is the case when the constant is negative:
.
When adding two vectors, add the corresponding components (this phrase comes from Wolfram Mathworld) of each vector. In other words, add the number on the same row to each other. For example, when adding 3u to (-2)v,
.
Apply the two rules for the four vector operations.
<h3>1.</h3>

Rewrite this vector as a linear combination of two unit vectors. The first component 69 will be the coefficient in front of the first unit vector, i. The second component 19 will be the coefficient in front of the second unit vector, j.
.
<h3>2.</h3>
.
<h3>3.</h3>
.
<h3>4.</h3>
.
All cross sections of spheres are “circles”
Answer:
The mean score for all students on the quiz was 86.
Step-by-step explanation:
This question is solved by a weighed average.
To find the weighed average, we add the values multiplied by their weight, and after that we divide by the sum of weights.
In this problem, we have that:
Mr. Spears had 20 students in his class with a mean score of 80.
So 80 has a weight of 20.
Mrs. Guyton's class of 30 students had a mean score of 90.
So 90 has a weight of 30.
Overall, what was the mean score for all students on the quiz

The mean score for all students on the quiz was 86.

is undefined if the argument

is negative, so you first need to require that

We're not done yet, though, because

still doesn't exist when

, so we remove this from the domain and we're left with

, or in interval notation,

To find the range, consider the limits of the function as you approach either endpoint of the domain.


Since

is positive everywhere, the range is