Answer:
Step-by-step explanation:
Given the function
y = (9x⁴ — 4x² + 6)⁴
We need to find the derivative of y with respect to x i.e. dy/dx.
So let u = 9x⁴—4x² + 6
Then y = u²,
Then, y is a function of u, y=f(u)
Also, u is a function of x, u = g(x)
In this case,
u = g(x) = 9x⁴—4x² + 6
So let differentiate this function y(x).
This is a function of a function
Then, we need to find u'(x)
u (x) = 9x⁴—4x² + 6
Then, u'(x) = 36x³ — 8x
Also we need to find y'(u)
Then, y = u²
y'(u) = 2u
Using function of a function formula
dy / dx = dy/du × du/dx
y'(x) = y'(u) × u'(x)
y'(x) = 2u × 36x³ — 8x
y'(x) = 2u(36x³ — 8x)
Since, u = 9x⁴—4x² + 6
Therefore,
y'(t) = 2(9x⁴—4x² + 6)(36x³ — 8x)
So,
dy/dx = 2(9x⁴—4x² + 6)(36x³ — 8x)
dy/dx = (18x⁴—8x² + 12)(36x³ — 8x)
Answer:
no.D) 18% is the answer of this question
1. You have that:
- The homeowner<span> want the length of the swimming pool to be 4 feet longer than its width.</span>
- He wants to surround it with a concrete walkway 3 feet wide.
- He can only afford 300 square feet of concrete for the walkway.
2. Therefore, the tota area is:
At=L2xW2
L2 is the lenght of the walkway (L2=L1+3+3⇒L2=(W1+4+6)⇒L2=W1+10).
W2 is the width of the walkway (W1+3+3⇒W2=W1+6)
3. The area of the walkway is:
A2=At-A1
A2=300 ft²
4. Therefore, you have that the width of the swimming pool is:
A2=(W1+10)(W1+6)-(W1+4)(W1)
300=(W1²+6W1+10W1+60)-(W1²+4W1)
W1²+16W1+60-W1²-4W1-300=0
12W1-240=0
W1=240/12
W1=20 ft
5. And the length is:
L1=W1+4
L1=20+4
L1=24 ft
Answer:
34.10
Step-by-step explanation:
Note:
16% = .16
Important:
What was the <u>old price</u> of the shoes
Solution:
40.60 x .16 = 6.946 [Round:6.50]
40.60-6.50= 34.10
Hence, the old price of the shoes is $34.10
Answer:
you put the points into the slope formula and solve : (y2-y1)/(x2-x1)
1. (4 -1) / (-2 -3)
3 / -5
2. (-1 -(-2)) / (5 - 0)
(-1 +2) / 5
1/ 5
Step-by-step explanation:
i hope this helped :)
