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Answer:
A)11
Step-by-step explanation:
These are matrices one dimensional with one column and 3 rows each.
-The product of the matrices is obtained by multiplying the correspond values and summing up;
![pq=\left[\begin{array}{ccc}3\\2\\-1\end{array}\right] \times\left[\begin{array}{ccc}5\\-1\\2\end{array}\right] \\\\\\\\=(3\times 5)+(2\times -1)+(-1\times 2)\\\\=15+-2+-2\\\\=11](https://tex.z-dn.net/?f=pq%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%5C%5C2%5C%5C-1%5Cend%7Barray%7D%5Cright%5D%20%5Ctimes%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C-1%5C%5C2%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5C%5C%5C%3D%283%5Ctimes%205%29%2B%282%5Ctimes%20-1%29%2B%28-1%5Ctimes%202%29%5C%5C%5C%5C%3D15%2B-2%2B-2%5C%5C%5C%5C%3D11)
Hence, the product of p and q is 11
Answer:
A
Step-by-step explanation:
according to tangent property
we get
Given that
Thus our equation is

substitute the given values

simplify multiplication:

simplify addition:

cancel 20 from both sides:

divide both sides by 2:

AN=CN
hence,CN=7
therefore our answer is A
Answer:
9 cans of soup and the 4 frozen dinners were purchased
Step-by-step explanation:
Let x represent the number of cans of soup purchased.
Let y represent the number of frozen dinners purchased.
Lincoln purchased a total of 13 cans of soup and frozen dinners. This means that
x + y = 13
Each can of soup has 250 mg of sodium and each frozen dinner has 550 mg of sodium. The 13 cans of soup and frozen dinners which he purchased collectively contain 4450 mg of sodium. This means that
250x + 550y = 4450 - - - - - - - - -1
Substituting x = 13 - y into equation 1, it becomes
250(13 - y) + 550y = 4450
3250 - 250y + 550y = 4450
- 250y + 550y = 4450 - 3250
300y = 1200
y = 1200/300
y = 4
Substituting y = 4 into x = 13 - y, it becomes
x = 13 - 4 = 9
Answer:
Last choice. d.
Step-by-step explanation:
We are given the first term and the sixth term.
The first term is
.
The sixth them is
.
Let's solve for the common ratio, r.

Divide both sides by -9:

Take the fifth root of both sides:


So the common ratio is 4.





