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3 years ago

100 points please help

Mathematics

2 answers:

yan [13]3 years ago

8 0

Answer:

Q30-A, C

Step-by-step explanation:

A) False, If the parallelogram VJHO was a rhombus, angle V wouldn't be supplementary to angle O

B) True, all opposite sides of a parallelogram parallel

C) False, the same reason as A

D) True, opposite sides of a parallelogram congruent

Q25-

angle BCA = 60 degrees

angle CAB = 27 degrees

angle BCT = 29 degrees

angle BMC = 58 degrees

torisob [31]3 years ago

3 0

Answer:

1. 2 and 4

2. BCA = 60

CAB = 27

BCT = 29

BMC = 58

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Check whether -2 and 2 are zeroes of the polynomial x+2

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Answer:

-2 is a zero of the polynomial. 2 is not a zero of the polynomial.

Step-by-step explanation:

A value of x is a zero of a polynomial if when it is substituted for x in the polynomial, it makes the polynomial evaluate to zero.

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3 years ago

Use integration by parts to find the integrals in Exercise.<br> x^3 ln x dx.

34kurt

Answer:

$\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C$ .

Step-by-step explanation:

We have been given an indefinite integral $\int \:x^3\:ln\:x\:dx$ . We are asked to find the value of the integral using integration by parts.

$\int\: u\text{dv}=uv-\int\: v\text{du}$

Let $u=\text{ln}(x)$ , $v'=x^3$ .

Now, we will find du and v as shown below:

$\frac{du}{dx}=\frac{d}{dx}(\text{ln}(x))$

$\frac{du}{dx}=\frac{1}{x}$

$du=\frac{1}{x}dx$

$v=\frac{x^{3+1}}{3+1}=\frac{x^{4}}{4}$

Upon substituting our values in integration by parts formula, we will get:

$\int \:x^3\:\text{ln}\:(x)\:dx=\text{ln}(x)*\frac{x^4}{4}-\int\: \frac{x^4}{4}*\frac{1}{x}dx$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\int\: \frac{x^3}{4}dx$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}\int\: x^3dx$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}*\frac{x^{3+1}}{3+1}+C$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}*\frac{x^4}{4}+C$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C$

Therefore, our required integral would be $\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C$ .

5 0

3 years ago