One could use the relation
![\mathbf a\cdot\mathbf b=\|\mathbf a\|\|\mathbf b\|\cos\theta](https://tex.z-dn.net/?f=%5Cmathbf%20a%5Ccdot%5Cmathbf%20b%3D%5C%7C%5Cmathbf%20a%5C%7C%5C%7C%5Cmathbf%20b%5C%7C%5Ccos%5Ctheta)
where
![\mathbf a,\mathbf b](https://tex.z-dn.net/?f=%5Cmathbf%20a%2C%5Cmathbf%20b)
are two vectors,
![\|\mathbf a\|](https://tex.z-dn.net/?f=%5C%7C%5Cmathbf%20a%5C%7C)
denotes the norm of that vector, and
![\theta](https://tex.z-dn.net/?f=%5Ctheta)
is the angle between the two vectors.
Then the two vectors will be parallel if
![\theta=0](https://tex.z-dn.net/?f=%5Ctheta%3D0)
or
![\pi](https://tex.z-dn.net/?f=%5Cpi)
.
Answer: 691
<u>Step-by-step explanation:</u>
There are 3 different ways to find the remainder. I am not sure which method you are supposed to use, so I will solve using all 3 methods.
Long Division:
<u> 10x³ + 24x² + 77x + 230 </u>
x - 3 ) 10x⁴ - 6x³ + 5x² - x + 1
- <u>(10x⁴ - 30x³) </u> ↓ ↓ ↓
24x³ + 5x² ↓ ↓
- <u>(24x³ - 72x²) </u> ↓ ↓
77x² - x ↓
- <u>(77x² - 231x) </u> ↓
230x + 1
- <u>(230x - 690)</u>
691 ← remainder
Synthetic Division:
x - 3 = 0 ⇒ x = 3
3 | 10 -6 5 -1 1
|<u> ↓ 30 72 231 690</u>
10 24 77 230 691 ← remainder
Remainder Theorem:
f(x) = 10x⁴ - 6x³ + 5x² - x + 1
f(3) = 10(3)⁴ - 6(3)³ + 5(3)² - (3) + 1
= 810 - 162 + 45 - 3 + 1
= 691
Answer:
E
Step-by-step explanation:
There are both red and green marbles in the jar. The answer cannot be A or D because that would mean that there's 1.14% or 1.04% green marbles. But we know that there's less than that in the and bag. And there are also red marbles. B and F do not work because there can't be a negative amount of marbles in the bag.
By changing the sign of x coordinates the point would be reflected along "y" axis.
Step-by-step explanation:
Let us plot a hypothetical point (x, y) where both x and y are positive integer. Since both the number sets are positive integer, hence the points lie in 1st quadrant.
Now we change the sign of the x coordinates i.e. x=-x
The new coordinates are (-x, y)
since x is negative and y is positive hence this set would lie in 2nd quadrant.
Comparing the above two pointset (x, y) and (-x, y)
We find that 2nd set is the reflection of the 1st set along y axis (“y” remains constant and sign of “x” coordinates change)
Hence if the original point is in 1st quadrant then the after changing the sign points would lie in 2nd quadrant and vice-versa.
Similarly, if both the values of “x” and “y” are negative i.e. points lie in 3rd quadrant, then the points would lie in 4th quadrant after changing and vice-versa
The answer is -7. Adding two negative make a negative then u add like u regularly would