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3 years ago

(8х + 2)(7х^2 + 6x - 7) Find the product

Mathematics

1 answer:

igomit [66]3 years ago

4 0

The product is 56x^3+62x^2-44x-14

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Add/sub the polynomial (1-8x^2+3x)+(1+6x^2+x)

Scorpion4ik [409]

Answer:

-2x² + 4x + 2

Step-by-step explanation:

(1-8x²+3x)+(1+6x²+x)

= 1 - 8x² + 3x + 1 + 6x² + x

= -2x² + 4x + 2 [-2(x² - 2 - 1)]

3 0

3 years ago

Use the drop down menus to complete the statements about the function p(x)=x(x-1)+1

soldi70 [24.7K]

Answer:

equal -1 plus 1 equals what

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2 years ago

A chi-square test for independence is being used to evaluate the relationship between two variables, one of which is classified

AysviL [449]

Answer:

Degrees of freedom for independence in chi-square statistic

ν = ( r-1) (s-1) =6

Step-by-step explanation:

Explanation:-

Given data chi-square test for independence is being used to evaluate the relationship between two variables

Given "A" is classified into 3 categories

Second 'B' is classified into 4 categories

In this chi-square test, we test if two attributes A and B under consideration are independent or not

We will assume that

Null Hypothesis : H₀: The two variables are independent

Degrees of freedom in chi-square test for independence

ν = ( r-1) (s-1)

Given data 'r' = 3 and 's' = 4

Degrees of freedom for independence

ν = ( r-1) (s-1) = ( 3-1) ( 4-1) = 2×3 =6

Test statistic

χ ² = ∑ $\frac{(O-E)^{2} }{E}$

4 0

3 years ago

Integration questions .

dlinn [17]

$\\\\\ \textbf{a)}\\\\~~~\displaystyle \int (6x- \sin 3x) ~ dx\\\\=6\displaystyle \int x ~ dx - \displaystyle \int \sin 3x ~ dx\\\\=6 \cdot \dfrac{x^2}2 - \dfrac 13 (- \cos 3x) +C~~~~~~~~~~~;\left[\displaystyle \int x^n~ dx = \dfrac{x^{n+1}}{n+1}+C,~~~n \neq -1\right]\\\\ =3x^2 +\dfrac{\cos 3x}3 +C~~~~~~~~~~~~~~~~~~~~;\left[\displaystyle \int \sin (mx) ~dx = -\dfrac 1m ~ (\cos mx)+C \right]\\$

$\textbf{b)}\\\\~~~~\displaystyle \int(3e^{-2x} +\cos (0.5 x)) dx\\\\=3\displaystyle \int e^{-2x} ~dx+ \displaystyle \int \cos(0.5 x) ~dx\\\\\\=-\dfrac 32 e^{-2x} + \dfrac 1{0.5} \sin (0.5 x) +C~~~~~~~~~~~~~~;\left[\displaystyle \int e^{mx}~dx = \dfrac 1m e^{mx} +C \right]\\\\\\=-\dfrac 32 e^{-2x} + 2 \sin(0.5 x) +C~~~~~~~~~~~~~~~~~;\left[\displaystyle \int \cos(mx)~ dx = \dfrac 1m \sin(mx) +C\right]\\\\\\=-1.5e^{-2x} +2\sin(0.5x) +C$

$\textbf{a)}\\\\y = \displaystyle \int \cos(x+5) ~ dx\\\\\text{Let,}\\\\~~~~~~~u = x+5\\\\\implies \dfrac{du}{dx} = 1+0~~~~~~;[\text{Differentiate both sides.}]\\\\\implies \dfrac{du}{dx} = 1\\\\\implies du = dx\\\\\text{Now,}\\\\y= \displaystyle \int \cos u ~ du\\\\~~~= \sin u +C\\\\~~~=\sin(x+5) + C$

$\textbf{b)}\\\\y = \displaystyle \int 2(5x-3)^4 dx\\\\\text{Let,}\\~~~~~~~~u = 5x-3\\\\\implies \dfrac{du}{dx} = 5~~~~~~~~~~;[\text{Differentiate both sides}]\\\\\implies dx = \dfrac{du}5\\\\\text{Now,}\\\\y = 2\cdot \dfrac 1 5 \displaystyle \int u^4 ~ du\\\\\\~~=\dfrac 25 \cdot \dfrac{u^{4+1}}{4+1} +C\\\\\\~~=\dfrac 25 \cdot \dfrac{u^5}5+C\\\\\\~~=\dfrac{2u^5}{25}+C\\\\\\~~=\dfrac{2(5x-3)^5}{25}+C$

$\textbf{a)}\\\\y = \displaystyle \int xe^{3x} dx\\\\\text{We know that,}\\\\ \displaystyle \int (uv) ~dx = u \displaystyle \int v ~ dx - \displaystyle \int \left[ \dfrac{du}{dx} \displaystyle \int ~ v ~ dx \right]~ dx\\\\\text{Let}, u =x~ \text{and}~ v=e^{3x} .\\\\y= \displaystyle \int xe^{3x} ~dx\\\\\\~~= x\displaystyle \int e^{3x} ~ dx - \displaystyle \int \left[\dfrac{d}{dx}(x) \displaystyle \int e^{3x}~ dx \right]~ dx\\\\\\$

$=x\displaystyle \int e^{3x}~ dx - \displaystyle \dfrac 13 \int \left(e^{3x} \right)~ dx\\\\\\=\dfrac{xe^{3x}}3 - \dfrac 13 \cdot \dfrac{ e^{3x}}3+C\\\\\\= \dfrac{xe^{3x}}{3}- \dfrac{e^{3x}}{9}+C\\\\\\=\dfrac{3xe^{3x}}{9}- \dfrac{e^{3x}}9 + C\\\\\\= \dfrac 19e^{3x}(3x-1)+C$

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8 0

2 years ago

I need help pls❗❗❗Its complicated

mel-nik [20]

Answer:

The answer to 36 is -18

Step-by-step explanation:

x=-18 because 3×(-6)=(-6×-6) cancels

3 0

3 years ago