It's not clear to me what the given interval is supposed to be, so I'll use a generic one, [a, b] with a < b.
The average acceleration of the particle over this interval is given by the average rate of change of v(t),
![a_{\rm ave} = \dfrac{v(b) - v(a)}{b - a} = \boxed{\dfrac{(b^{0.7}+4b)\cos(b^2) - (a^{0.7}+4a)\cos(a^2)}{b - a}}](https://tex.z-dn.net/?f=a_%7B%5Crm%20ave%7D%20%3D%20%5Cdfrac%7Bv%28b%29%20-%20v%28a%29%7D%7Bb%20-%20a%7D%20%3D%20%5Cboxed%7B%5Cdfrac%7B%28b%5E%7B0.7%7D%2B4b%29%5Ccos%28b%5E2%29%20-%20%28a%5E%7B0.7%7D%2B4a%29%5Ccos%28a%5E2%29%7D%7Bb%20-%20a%7D%7D)
FACTORISATION IS BASICALLY FINDING LIKE-TERMS.
1. FIND THE HCF OF 8. HERE, THE HCF IS 8. ALSO, CHECK ALL THE LIKE-TERMS. LIKE-TERMS ARE OUTSIDE THE BRACKET AND UNLIKE TERMS ARE INSIDE.
2. ADD THE TERMS IN THE BRACKET AND THE TERMS OUTSIDE THE BRACKET FROM STEP 1.
8x + 8y + rx + ry
1. 8 (x + y) + r (x + y)
2. (x +y) (8+r)
Answer:
Math isnt my strongest subjuect but i believe it it right.
Not sure about the statement
movement down 5 units
move right 2 units
Answer:
Step-by-step explanation:
![R = \frac{ {x}^{2} }{y} \\ \\ = \frac{ {(3.8 \times {10}^{5}) }^{2} }{5.9 \times {10}^{4} } \\ \\ = \frac{14.44 \times {10}^{10} }{5.9 \times {10}^{4} } \\ \\ = 2.44745763 \times {10}^{10 - 4} \\ = 2.44745763 \times {10}^{6}](https://tex.z-dn.net/?f=R%20%3D%20%20%5Cfrac%7B%20%7Bx%7D%5E%7B2%7D%20%7D%7By%7D%20%20%5C%5C%20%20%5C%5C%20%20%3D%20%20%5Cfrac%7B%20%7B%283.8%20%5Ctimes%20%20%7B10%7D%5E%7B5%7D%29%20%7D%5E%7B2%7D%20%7D%7B5.9%20%5Ctimes%20%20%7B10%7D%5E%7B4%7D%20%7D%20%20%5C%5C%20%20%5C%5C%20%20%3D%20%20%5Cfrac%7B14.44%20%5Ctimes%20%20%7B10%7D%5E%7B10%7D%20%7D%7B5.9%20%5Ctimes%20%20%7B10%7D%5E%7B4%7D%20%7D%20%20%5C%5C%20%20%5C%5C%20%20%3D%202.44745763%20%5Ctimes%20%20%7B10%7D%5E%7B10%20-%204%7D%20%20%5C%5C%20%3D%202.44745763%20%5Ctimes%20%20%7B10%7D%5E%7B6%7D%20)