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3 years ago

7

Use the Binomial Theorem Which of the following is equivalent to the expression (m + 3)4

1 answer:

kondor19780726 [428]3 years ago

3 0

Answer:

Step-by-step explanation:

(x+a)^n=nc0x^n+nc1x^{n-1}a+nc2x^{n-2}a^2+...+ncna^n

(x+3)^4=4c0m^4+4c1m^3(3)+4c2m^2(3)^2+4c3m(3)^3+4c4(3)^4

4c0=1

4c1=4/1=4

4c2=(4×3}/(2×1)=6

4c3=(4×3×2)/(3×2×1)=4

4c4=(4×3×2×1)/(4×3×2×1)=1

(m+3)^4=m^4+4m³(3)+6m²(3)²+4m(3)³+3^4

=m^4+12m³+54m²+108m+81

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2 years ago

Solve the following equation:<br> 9n² - n - 2/3 = 0

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Step-by-step explanation:

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9n² - n - $\frac{2}{3}$ = 0 ← multiply through by 3 to clear the fraction

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2 years ago

Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.

Levart [38]

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is $Q(x,y)$ is predicate that $x$ then $x^{2}$ . where $x,y$ are rational numbers.

a)

when $x=-2, y=1$

Here $-2$ that is $x$ satisfied. Then

$(-2)^{2}$

$4$ this is wrong. since $4>1$

That is $x^{2}$ $>y^{2}$ Thus $Q(x,y)$ $=Q(-2,1)$ is false.

b)

Assume $Q(x,y)=Q(-5,2)$ .

That is $x=-5, y=2$

Here $-5$ that is $x$ this condition is satisfied.

Then

$(-5)^{2}$

$25$ this is not true. since $25>4$ .

This is similar to the truth value of part (a).

Since in both $x$ satisfied and $x^{2} >y^{2}$ for both the points.

c)

if $Q(x,y)=Q(3,8)$ that is $x=3$ and $y=8$

Here $3$ this satisfies the condition $x$ .

Then $3^{2}$

$9$ This also satisfies the condition $x^{2}$ .

Hence $Q(3,8)$ exists and it is true.

d)

Assume $Q(x,y)=Q(9,10)$

Here $9$ satisfies the condition $x$

Then $9^{2}$

$81$ satisfies the condition $x^{2}$ .

Thus, $Q(9,10)$ point exists and it is true. This satisfies the same values as in part (c)

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