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mezya [45]
3 years ago
13

PLEASE HELP ME ASAP I NEED HELP ITS DUE AT 11:59

Mathematics
1 answer:
ivanzaharov [21]3 years ago
8 0

Answer:

When -8 is x then y = -29

When -4 is x then y = -17

When 4 is x then y = 7

When 6 is x then y = 13

Step-by-step explanation:

You simply substitute the numbers on the x row for x in the equation, then you solve the problem. (3 x -8) - 5, ect.

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Evaluate the following integrals: 1. Z x 4 ln x dx 2. Z arcsin y dy 3. Z e −θ cos(3θ) dθ 4. Z 1 0 x 3 √ 4 + x 2 dx 5. Z π/8 0 co
Zigmanuir [339]

Answer:

The integrals was calculated.

Step-by-step explanation:

We calculate integrals, and we get:

1) ∫ x^4 ln(x) dx=\frac{x^5 · ln(x)}{5} - \frac{x^5}{25}

2) ∫ arcsin(y) dy= y arcsin(y)+\sqrt{1-y²}

3) ∫ e^{-θ} cos(3θ) dθ = \frac{e^{-θ} ( 3sin(3θ)-cos(3θ) )}{10}

4) \int\limits^1_0 {x^3 · \sqrt{4+x^2} } \, dx = \frac{x²(x²+4)^{3/2}}{5} - \frac{8(x²+4)^{3/2}}{15} = \frac{64}{15} - \frac{5^{3/2}}{3}

5)  \int\limits^{π/8}_0 {cos^4 (2x) } \, dx =\frac{sin(8x} + 8sin(4x)+24x}{6}=

=\frac{3π+8}{64}

6)  ∫ sin^3 (x) dx = \frac{cos^3 (x)}{3} - cos x

7) ∫ sec^4 (x) tan^3 (x) dx = \frac{tan^6(x)}{6} + \frac{tan^4(x)}{4}

8)  ∫ tan^5 (x) sec(x)  dx = \frac{sec^5 (x)}{5} -\frac{2sec^3 (x)}{3}+ sec x

6 0
3 years ago
Solve: 3x - 7 - 2x + 5 = 6 A) -8 B) - 18 5 C) 4 D) 8
seropon [69]

Answer:

D) 8

Step-by-step explanation:

3x - 7 - 2x + 5 = 6

3x - 2x  = 6 + 7 - 5

  x   = 8

Note: 6 + 7 -5 =  13 -5 = 8

7 0
3 years ago
Which is the graph of f(x)=sqrt x?
Svet_ta [14]

Answer:

B

Step-by-step explanation:

a parabola would be x2 since it is square rooted it would be a curved so the answer would be the second graph

3 0
3 years ago
If you take the square root of every whole number from 1 through 100, how
Lady_Fox [76]
Only numbers 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100 will come out to be whole numbers.

Meaning, that 10 are whole numbers and the other 90 are not.
8 0
3 years ago
Which equation is not a polynomial.
babymother [125]
A. -sqrt(2)

Explanation:
there are no degrees, etc
7 0
3 years ago
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