Answer: The knights were responsible for protecting the people.
Answer:
y= 5x+13
Step-by-step explanation:
<u>Slope- intercept form</u>
y= mx +b, where m is the gradient and b is the y-intercept.
Parallel lines have the same gradient.
y= 5x +4
Gradient of given line= 5
Thus, gradient of line= 5
Subst. m=5 into the equation.
y= 5x +b
To find the value of b, substitute a coordinate
When x= -2, y=3,
3= 5(-2) +b
3= -10 +b
b= 3 +10 <em>(</em><em>+</em><em>1</em><em>0</em><em> </em><em>on</em><em> </em><em>both</em><em> </em><em>sides</em><em>)</em>
b= 13
Thus, the equation of the line is y= 5x +13.
Answer:
840
Step-by-step explanation:
Last year, 800 students attended the career fair at West High School.
And, this year the number of students who attended the career fair increased by 5%.
We are asked how many students attended the career fair at West High School this year.
Therefore, the number of students attended the career fair at West High School will be ![800[1+\frac{5}{100} ] =800 \times 1.05 = 840](https://tex.z-dn.net/?f=800%5B1%2B%5Cfrac%7B5%7D%7B100%7D%20%5D%20%20%3D800%20%5Ctimes%201.05%20%3D%20840)
. (Answer)
-4(n - 6) = 12
n - 6 = -12/4 divide both sides by -4
n - 6 = -3
n = -3 + 6 add both sides by 6
n = 3
hope that helps, God bless!
Answer:
- r(0) = <0, 100> . . . . . . . .meters
- r'(0) = <7.071, 7.071> . . . . meters per second
Step-by-step explanation:
<u>Initial Position</u>
The problem statement tells us we're measuring position from the ground at the base of the building where the projectile was launched. The initial horizontal position is presumed to be zero. The initial vertical position is said to be 100 meters from the ground, so (in meters) ...
r(0) = <0, 100>
<u>Initial Velocity</u>
The velocity vector resolves into components in the horizontal direction and the vertical direction. For angle α from the horizontal, the horizontal component of velocity is v₁·cos(α), and the vertical component is v₁·sin(α). For v₁ = 10 m/s and α = π/4, the initial velocity vector (in m/s) is ...
r'(0) = <10·cos(π/4), 10·sin(π/4)>
r'(0) ≈ <7.071, 7.071>
