2 triangles make up a rhombus
For question number 1:The plot H = H(t) is the parabola and it reaches its maximum in the moment when exactly at midpoint between the roots t = 0 and t = 23. At that moment t = 23/2 or 11.5 seconds.
For question number 2:To find the maximal height, just simply substitute t = 11.5 into the quadratic equation. The answer would be 22.9.
For question number 3:H(t) = 0, or, which is the same as -16t^2 + 368t = 0.Factor the left side to get -16*t*(t - 23) = 0.t = 0, relates to the very start of the process, when the ash started its way up.The other root is t = 23 seconds, and it is precisely the time moment when the bit of ash will go back to the ground.
The picture is not clear. let me assume
y = (x^4)ln(x^3)
product rule :
d f(x)g(x) = f(x) dg(x) + g(x) df(x)
dy/dx = (x^4)d[ln(x^3)/dx] + d[(x^4)/dx] ln(x^3)
= (x^4)d[ln(x^3)/dx] + 4(x^3) ln(x^3)
look at d[ln(x^3)/dx]
d[ln(x^3)/dx]
= d[ln(x^3)/dx][d(x^3)/d(x^3)]
= d[ln(x^3)/d(x^3)][d(x^3)/dx]
= [1/(x^3)][3x^2] = 3/x
... chain rule (in detail)
end up with
dy/dx = (x^4)[3/x] + 4(x^3) ln(x^3)
= x^3[3 + 4ln(x^3)]
Answer: 3 4/9 and 4 19/29
Step-by-step explanation: Mixed fractions can be converted into improper fractions. Example: 3 1/2=
3×2+1= 7/2
So, 3 4/9= 3×9+4= 31/9
4 19/29= 4×29+19= 135/29
PLEASE RATE 5 STARS AND VOTE AS BRAINLIEST:)
(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)
Answer:
x ≥-3
Step-by-step explanation:
7 − 9x − (x + 12) less than or equal to 25
7 − 9x − (x + 12) ≤ 25
Distribute the minus sign
7 − 9x − x - 12 ≤ 25
Combine like terms
− 10x −5 ≤ 25
Add 5 to each side
− 10x −5+5 ≤ 25+5
− 10x ≤ 30
Divide by -10. Remember that this flips the inequality
− 10x/-10 ≥30/-10
x ≥-3
