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tatuchka [14]
3 years ago
12

Select all the expressions that are equivalent to y'.

Mathematics
1 answer:
aleksley [76]3 years ago
3 0

Answer:

A, B, D

Step-by-step explanation:

A) y squared times y squared you just add the exponents

B) y^6/y^2 write it out as (y)(y)(y)(y)(y)(y)/(y)(y) then u cancel out two "y" from numerator and denominator and is left with y^4

C) is just 0 so its wrong

D) is the same process as B just cancel out 5 "y" from numerator and denominator and is left with y^4

E) is incorrect bc you need to add the exponents and you will get y^5 which is not equal to y^4 so yeah its wrong

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Answer:

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Anna007 [38]

Answer:

the answer is A

Step-by-step explanation:

8 0
3 years ago
100 points , please help. I am not sure if I did this correct if anyone can double-check me thanks!
Nookie1986 [14]

Step-by-step explanation:

\lim_{n \to \infty} \sum\limits_{k=1}^{n}f(x_{k}) \Delta x = \int\limits^a_b {f(x)} \, dx \\where\ \Delta x = \frac{b-a}{n} \ and\ x_{k}=a+\Delta x \times k

In this case we have:

Δx = 3/n

b − a = 3

a = 1

b = 4

So the integral is:

∫₁⁴ √x dx

To evaluate the integral, we write the radical as an exponent.

∫₁⁴ x^½ dx

= ⅔ x^³/₂ + C |₁⁴

= (⅔ 4^³/₂ + C) − (⅔ 1^³/₂ + C)

= ⅔ (8) + C − ⅔ − C

= 14/3

If ∫₁⁴ f(x) dx = e⁴ − e, then:

∫₁⁴ (2f(x) − 1) dx

= 2 ∫₁⁴ f(x) dx − ∫₁⁴ dx

= 2 (e⁴ − e) − (x + C) |₁⁴

= 2e⁴ − 2e − 3

∫ sec²(x/k) dx

k ∫ 1/k sec²(x/k) dx

k tan(x/k) + C

Evaluating between x=0 and x=π/2:

k tan(π/(2k)) + C − (k tan(0) + C)

k tan(π/(2k))

Setting this equal to k:

k tan(π/(2k)) = k

tan(π/(2k)) = 1

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1/(2k) = 1/4

2k = 4

k = 2

8 0
3 years ago
97+(-27)+(-89)+(-54)
RUDIKE [14]

Step 1:

97+(-27)+(-89)+(-54)

Step 2:

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Info:

(+) ×(-)=-

Step 3:

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Step 4:

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sergiy2304 [10]
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