Answer:
A
Step-by-step explanation:
![\left[\begin{array}{ccc}12 &-6\\1 &-10 \\\end{array}\right] +\left[\begin{array}{ccc}-2&14\\5&15\\\end{array}\right] =\left[\begin{array}{ccc}12-2&-6+14\\1+5&-10+15\\\end{array}\right] =\left[\begin{array}{ccc}10&8\\6&5\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%20%26-6%5C%5C1%20%26-10%20%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%2614%5C%5C5%2615%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12-2%26-6%2B14%5C%5C1%2B5%26-10%2B15%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D10%268%5C%5C6%265%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Answer:
we need the graph to answer the question
Step-by-step explanation:
we wont know much he makes each week without the graph
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Answer:
C
Step-by-step explanation:
Given 2 secants intersect a circle from a point outside the circle, then
The product of the external part and the entire part of one secant is equal to the product of the external part and the entire part of the other secant, that is
x(x + 10 + x) = 6(6 + 10 + x)
x(2x + 10) = 6(16 + x) ← distribute parenthesis on both sides
2x² + 10x = 96 + 6x ← subtract 96 + 6x from both sides
2x² + 4x - 96 = 0 ← in standard form
Divide through by 2
x² + 2x - 48 = 0 ← factor the left side
(x + 8)(x - 6) = 0
Equate each factor to zero and solve for x
x + 8 = 0 ⇒ x = - 8
x - 6 = 0 ⇒ x = 6
However x > 0 ⇒ x = 6 → C
This is a strange question, and f(x) may not even exist. Why do I say that? Well..
[1] We know that f(a+b) = f(a) + f(b). Therefore, f(0+0) = f(0) + f(0). In other words, f(0) = f(0) + f(0). Subtracting, we see, f(0) - f(0) = f(0) or 0 = f(0).
[2] So, what's the problem? We found the answer, f(0) = 0, right? Maybe, but the second rule says that f(x) is always positive. However, f(0) = 0 is not positive!
Since there is a contradiction, we must either conclude that the single value f(0) does not exist, or that the entire function f(x) does not exist.
To fix this, we could instead say that "f(x) is always nonnegative" and then we would be safe.
