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MrRa [10]
3 years ago
8

3. What are the values of x and y in each figure? 35 77 96

Mathematics
1 answer:
jeka943 years ago
7 0
96 I think correct me If I’m wrong
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What is the derivative of x^3
wolverine [178]
The derivative here is what you get when you bring the exponent down as the coefficient (i.e. the 3) and subtract the exponent by 1. In this case, the answer will be 3x^2.
8 0
3 years ago
What is the lim as x approaches pi [integral 1+tan t from pi to x]/(pi sin x)
MissTica

Apply l'Hopital's rule:

\displaystyle\lim_{x\to\pi}\frac{\displaystyle\int_\pi^x (1+\tan(t))\,\mathrm dt}{\pi\sin(x)}=\lim_{x\to\pi}\frac{1+\tan(x)}{\pi\cos(x)}=\frac{1+\tan(\pi)}{\pi\cos(\pi)}=\boxed{-\frac1\pi}

where

\displaystyle\frac{\mathrm d}{\mathrm dx}\left[\int_\pi^x(1+\tan(t))\,\mathrm dt\right]=1+\tan(x)

follows from the fundamental theorem of calculus.

3 0
3 years ago
For the function g(x)=−8x−8, evaluate g(−4).
Novosadov [1.4K]

24

Step-by-step explanation:

Given in pic. have a good day!!

8 0
3 years ago
2 1/3 times 3 1/2<br> As a mixed number
Sedbober [7]

Answer:

8 1/6

Step-by-step explanation:

8 1/6 is the answer. Hope this helps!

8 0
3 years ago
Read 2 more answers
Help me with these please​
olchik [2.2K]

Step-by-step explanation:

(1) y = x e^(x²)

Take derivative with respect to x:

dy/dx = x (e^(x²) 2x) + e^(x²)

dy/dx = 2x² e^(x²) + e^(x²)

dy/dx = (2x² + 1) e^(x²)

Take derivative with respect to x again:

d²y/dx² = (2x² + 1) (e^(x²) 2x) + (4x) e^(x²)

d²y/dx² = (4x³ + 2x) e^(x²) + 4x e^(x²)

d²y/dx² = (4x³ + 6x) e^(x²)

Substitute:

d²y/dx² − 2x dy/dx − 4y

= (4x³ + 6x) e^(x²) − 2x (2x² + 1) e^(x²) − 4x e^(x²)

= 4x³ + 6x − 2x (2x² + 1) − 4x

= 4x³ + 6x − 4x³ − 2x − 4x

= 0

(2) y = sin⁻¹(√x)

sin y = √x

sin²y = x

Take derivative with respect to x:

2 sin y cos y dy/dx = 1

sin(2y) dy/dx = 1

dy/dx = csc(2y)

Take derivative with respect to x again:

d²y/dx² = -csc(2y) cot(2y) 2 dy/dx

d²y/dx² = -2 csc²(2y) cot(2y)

Substitute:

2x (1 − x) d²y/dx² + (1 − 2x) dy/dx

= 2 sin²y (1 − sin²y) (-2 csc²(2y) cot(2y)) + (1 − 2 sin²y) csc(2y)

Use power reduction formula:

= (1 − cos(2y)) (1 − ½ (1 − cos(2y))) (-2 csc²(2y) cot(2y)) + (1 − (1 − cos(2y))) csc(2y)

= (1 − cos(2y)) (1 − ½ + ½ cos(2y)) (-2 csc²(2y) cot(2y)) + cos(2y) csc(2y)

= (1 − cos(2y)) (½ + ½ cos(2y)) (-2 csc²(2y) cot(2y)) + cot(2y)

= (cos(2y) − 1) (1 + cos(2y)) csc²(2y) cot(2y) + cot(2y)

= (cos²(2y) − 1) csc²(2y) cot(2y) + cot(2y)

= -sin²(2y) csc²(2y) cot(2y) + cot(2y)

= -cot(2y) + cot(2y)

= 0

8 0
3 years ago
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