Ask question

Ask question

All categories

3 years ago

14

I REALLY NEED HELP PLS Use the Binomial Theorem to find the binomial expansion of the expression. (x-3)^5

1 answer:

Lyrx [107]3 years ago

8 0

$(x - 3)^{5}$
$(x - 3)(x - 3)(x - 3)(x - 3)(x - 3)$
$(x^{2} - 6x + 9)(x^{2} - 6x + 9)(x^{2} - 6x + 9)(x - 3) \\(x^{4} - 12x^{3} + 54x^{2} - 108x + 81)(x - 3) \\x^{5} - 15x^{4} + 90x^{3} - 270x^{2} + 189x - 324$

You might be interested in

Find the values for c and d

baherus [9]

Answer:

c=-5

d=1

Step-by-step explanation:

$(cy^2)(4y^d)=-20y^3$

I'm going to reorder the left-hand side. Multiplication is commutative.

$(4c)(y^2y^d)=-20y^3$

Since the bases are the same in $y^2y^d$ and the operation is multiplication, I'm going to add the exponents giving me:

$4cy^{2+d}=-20y^3$

So this implies we have two equations to solve:

$4c=-20$ and $2+d=3$

So the first equation can be solved by dividing both sides by 4 giving you $c=-5$ .

The second equation can be solved by subtracting 2 on both sides giving you $d=1$ .

6 0

3 years ago

I'm not sure what they mean by "relations that represents a function". I don't understand this.

sergeinik [125]

Means, for every 1 input, there is 1 coresponding output
normally x is input and y is output in form (x,y)
so just look for the option(s) that has/have every first number repeat with only 1 2nd number

A. -7 repeats with 5 and -3, not a function

B. no repeats of first number (6,-8) and (-5,-8) are fine because first numbers don't repeat, das is funciton

C. no repeats, function

D. no repeats, function

answer is B,C,D

4 0

3 years ago

ASAP<br> -WILL MARK BRIANIST

nalin [4]

83-8 =75

Therefore, 75 =b

6 0

3 years ago

azamat

Answer:

(x-7)^2+(y-8)^2=121

Step-by-step explanation:

4 0

2 years ago

i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

3 0

3 years ago