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3 years ago

Pls help me plssssssssss

Mathematics

1 answer:

MariettaO [177]3 years ago

7 0

Step-by-step explanation:

what type of help u want......

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Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

3 years ago

2x² - 5x+1 has roots alpha and beta. Find alpha⁴+beta⁴ without solving the equation.

shepuryov [24]

Answer:

Step-by-step explanation:

$\alpha+\beta=\dfrac{5}{2} \\\\\alpha*\beta=\dfrac{1}{2} \\\\\alpha^2+\beta^2=\dfrac{21}{4} \ (see\ previous\ post)\\\\(\alpha+\beta)^4=\dfrac{625}{16} \\\\=\alpha^4+\beta^4+4*\alpha^3*\beta+6*\alpha^2*\beta^2+4*\alpha*\beta^3\\\\=\alpha^4+\beta^4+4*(\alpha*\beta)(\alpha^2+\beta^2)+6*\alpha^2*\beta^2\\\\\alpha^4+\beta^4=(\alpha+\beta)^4-4*(\alpha*\beta)(\alpha^2+\beta^2)-6*\alpha^2*\beta^2\\\\= \dfrac{625}{16} -4*\dfrac{1}{2} *\dfrac{21}{4} -6*(\dfrac{1}{2})^2 \\\\$

$= \dfrac{625}{16}- \dfrac{168}{16}-\dfrac{24}{16}\\\\\\= \dfrac{433}{16}$

7 0

3 years ago

Solve for k in g=k-a.

Varvara68 [4.7K]

Answer:

k = g + a

Step-by-step explanation:

g = k - a ( add a to both sides )

g + a = k , that is

k = g + a

5 0

2 years ago

Plz plz help me I need your help plz help me I’m begging you so much plz plz help me I need your help plz help me

Paraphin [41]

I am pretty sure the answer is A.

5 0

3 years ago

Read 2 more answers

Write an equation of the line that passes through (1,2) and is perpendicular to the line y= 1/3x+4

zvonat [6]

Slope-intercept form: y = mx + b

(m is the slope, b is the y-intercept or the y value when x = 0 --> (0, y) or the point where the line crosses through the y-axis)

For lines to be perpendicular, their slopes have to be the negative reciprocal of each other. (Basically flip the sign +/- and the fraction(switch the numerator and the denominator))

For example:

Slope = 2 or $\frac{2}{1}$

Perpendicular line's slope = $-\frac{1}{2}$ (flip the sign from + to -, and flip the fraction)

Slope = $-\frac{3}{4}$

Perpendicular line's slope = $\frac{4}{3}$ (flip the sign from - to +, and flip the fraction)

y = 1/3x + 4 The slope is 1/3, so the perpendicular line's slope is $-\frac{3}{1}$ or -3.

Now that you know the slope, substitute/plug it into the equation:

y = mx + b

y = -3x + b To find b, plug in the point (1, 2) into the equation, then isolate/get the variable "b" by itself

2= -3(1) + b Add 3 on both sides to get "b" by itself

2 + 3 = -3 + 3 + b

5 = b

y = -3x + 5

5 0

3 years ago