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(6.508x10^-3)x(8.837x10^-3)/(2.446x10^-15) pls help

\(\frac{(6.508\times10^{-3})\times(8.837\times 10^{-3})}{(2.446\times10^{-15}) }\\ =\frac{6.508\times8.837\times 10^{-3+-3--15})}{2.446}\\ =\frac{6.508\times8.837}{2.446}\times 10^{-3-3+15}\\ =\frac{6.508\times8.837}{2.446}\times 10^{9}\\ \approx 23.12345\times 10^{9}\\ \approx 2.312345\times 10^1\times 10^{9}\\ \approx 2.312345\times 10^{10}\\ \)

write csc theta and cot theta in terms of sin theta and cos theta.

\(csc\;\theta = \frac{1}{sin\;\theta}\\~\\ \\ Cot\;\theta = \frac{cos\;\theta}{\sin\;\theta}\)

Yes that is correct :)

i)  Show that for all positiveintegers n,

\(x[(1+x)^{n-1}+(1+x)^{n-2}+\dots (1+x)+1]=(1+x)^n-1\qquad (1)\\\)

I can do this first bit no worries but then part 2 is

Hence show that for   \(1\le k\le n\)

\(\begin{pmatrix}n-1 \\k-1\end{pmatrix}+ \begin{pmatrix}n-2 \\k-1\end{pmatrix}+ \begin{pmatrix}n-3 \\k-1\end{pmatrix}+\dots + \begin{pmatrix}k-1 \\k-1\end{pmatrix}= \begin{pmatrix}n \\k\end{pmatrix}\)

I don't know how to do this second bit.  Can someone help me please?

Ok what you need to see is that is the coefficient of x^k  in the right hand side of equation 1 is   nCk

So I need to find the coefficient of x^k in the left hand side.

Those two coefficients mucs be equal

Looking at equation (1)

\(RHS\\ = (1+x)^n-1\\ =\begin{pmatrix}n\\0\end{pmatrix} +\begin{pmatrix}n\\1\end{pmatrix}x +\begin{pmatrix}n\\2\end{pmatrix}x^2\dots +\begin{pmatrix}n\\k\end{pmatrix}x^k\dots +\begin{pmatrix}n\\n\end{pmatrix}x^n-1\\ \therefore \mbox{The coefficient of }x^k \;\;is\;\;\begin{pmatrix} n\\k \end{pmatrix}\)

\(LHS=x[(1+x)^{n-1}+(1+x)^{n-2}+\dots (1+x)+1]\\ \mbox{I want to find the coefficient of }x^k\\ \mbox{So I am only interested in }x^{k-1} \;\;\mbox{terms in the binomial expansions}\\ x\left [\begin{pmatrix} n-1\\k-1 \end{pmatrix}x^{k-1}+\begin{pmatrix} n-2\\k-1 \end{pmatrix}x^{k-1}+\dots +\begin{pmatrix} k-1\\k-1 \end{pmatrix}x^{k-1} \right]\\ =\left [\begin{pmatrix} n-1\\k-1 \end{pmatrix}+\begin{pmatrix} n-2\\k-1 \end{pmatrix}+\dots +\begin{pmatrix} k-1\\k-1 \end{pmatrix}\right]x^k\\ \)

Equating coefficients we have

\(\begin{pmatrix} n-1\\k-1 \end{pmatrix}+\begin{pmatrix} n-2\\k-1 \end{pmatrix}+\dots +\begin{pmatrix} k-1\\k-1 \end{pmatrix}= \begin{pmatrix} n\\k \end{pmatrix} \)

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So glad someone asked this,I was stuck on this question for my homework for a hour.

Thank you so much! You have no idea how much this means to me! It is super helpful!!!! Thanks! 😊😊😊

1. If V= 12R/(r+R) make R the subject of the equation.

Solve for R:
V = (12 R)/(r+R)

V = (12 R)/(r+R) is equivalent to (12 R)/(r+R) = V:
(12 R)/(r+R) = V

Multiply both sides by r+R:
12 R = V (r+R)

Expand out terms of the right hand side:
12 R = r V+R V

Subtract R V from both sides:
R (12-V) = r V

Divide both sides by 12-V:
Answer: |R = (r V)/(12-V)

2a. Jane, Maria and Ben each have a collection of marbles. Jane has 15 more marbles than Ben, and Maria has 2 times as many marbles as Ben. All together, they have 95 marbles. Find how many marbles Maria has.

Let Ben's marbles =B

Jane has =B+15 marbles

Maria has =2B marbles

B + B + 15 + 2B =95

4B=95 - 15

4B =80 divide both sides by 4

B=20 Ben's marbles

20+15 =35 Jane's marbles

2 x 20 =40 Maria's marbles.

2b. Dave sold 40 tickets for a concert. He sold x tickets at $2 each and y tickets at $3 each. He collected $88. Write down two equations connecting x and y. Solve these two equations to find how many of each kind of ticket he sold

x + y =40

2x + 3y =88

Solve the following system:
{x+y = 40 | (equation 1)
2 x+3 y = 88 | (equation 2)
Swap equation 1 with equation 2:
{2 x+3 y = 88 | (equation 1)
x+y = 40 | (equation 2)
Subtract 1/2 × (equation 1) from equation 2:
{2 x+3 y = 88 | (equation 1)
0 x-y/2 = -4 | (equation 2)
Multiply equation 2 by -2:
{2 x+3 y = 88 | (equation 1)
0 x+y = 8 | (equation 2)
Subtract 3 × (equation 2) from equation 1:
{2 x+0 y = 64 | (equation 1)
0 x+y = 8 | (equation 2)
Divide equation 1 by 2:
{x+0 y = 32 | (equation 1)
0 x+y = 8 | (equation 2)
Collect results:
Answer: |x = 32                         and                           y = 8

2c. A rectangle has length (x+5) cm and width of (x-2) cm. Its area is 60 cm^2. Write a quadratic equation, and solve it to fins the length and the width of this rectangle.

Solve for x:
(x-2) (x+5) = 60

Expand out terms of the left hand side:
x^2+3 x-10 = 60

Add 10 to both sides:
x^2+3 x = 70

Add 9/4 to both sides:
x^2+3 x+9/4 = 289/4

Write the left hand side as a square:
(x+3/2)^2 = 289/4

Take the square root of both sides:
x+3/2 = 17/2 or x+3/2 = -17/2

Subtract 3/2 from both sides:
x = 7 or x+3/2 = -17/2

Subtract 3/2 from both sides:
Answer: |x = 7

For the Logarithm Question:

  Assuming real numbers, square roots will be defined when the value inside the square root sign is zero or positive:

     sqrt(x + 1) is defined for all values:  x + 1 >= 0     --->     x >= -1

     sqrt(1 - x) is defined for all values:  1 - x >= 0     --->     -x >= -1     --->     x <= 1

     sqrt(x) is defined for all values:  x >= 0     --->     but a denominator can't be zero     --->     x > 0

          Putting these restrictions together:     0 < x <= 1

Polynomial Question #1:

   The degreee of a polynomial is the degree of its highest degree term.

   f(x) is quadratic   --->   its highest degree term is 2

   g(x) is cubic   --->   its highest degree terms is 3

   [ f(x) ]³  has a highest degree of  [ x² ]³  =  x⁶    

        --->   degree is 6 (with a coefficient of 1)  (but it can also have terms of degree 5, degree 4, etc.)

   [ g(x) ]²  has a highest degree of  [ x³ ]²  =  x⁶   

          --->   degree is 6 (with a coefficient of 1) (but it can also have terms of degree 5, degree 4, etc.)

   Since the coefficient of the  x⁶  term of both the [ f(x) ]3  term and the  [ g(x) ]²  term is 1, these terms will cancel      

   under subtraction, leaving a possible  x⁵  term.

--->     The maximum possible degree of the answer is 5.

Polynomial Question #2:

 If  f(x)  =  x⁷  and  g(x)  =  -x⁷,  then  f(x) + g(x)  =  x⁷ + -x⁷   =  0, which has a degree of 0, the minimum possible degree.

If  f(x)  =  x⁷  and  g(x)  =  x⁷,  then  f(x) + g(x)  =  x⁷ + x⁷   =  2x⁷,  which has a degree of 7, the maximum possible degree.

Multiplying these two answers:  0 x 7 = 0.

When Jerry was 4, his sister was 2. That means they are separated by 2 years only.

That means if Jerry is 100, then his sister is 98 years old.

2x + xy = xz + 3 Solve for x

Solve for x:
2 x+x y = 3+x z

Collect in terms of x:
x (y+2) = 3+x z

Subtract x z from both sides:
x (2+y-z) = 3

Divide both sides by 2+y-z:
Answer: |x = 3/(2+y-z)

1/x + 1/2 = y Solve for x

Solve for x:
1/2+1/x = y

Bring 1/2+1/x together using the common denominator 2 x:
(x+2)/(2 x) = y

Multiply both sides by 2 x:
x+2 = 2 x y

Subtract 2+2 x y from both sides:
x (1-2 y) = -2

Divide both sides by 1-2 y:
Answer: |x = -2/(1-2 y)

Since you are multiplying 86 by 97 right after you are dividing 86 by 97, and multiplication and division are opposite opertations, they cancel each other out, and your answer is 86. 

The domain of a function f(x) is the values that x can be. 

If you want to know how to find the domain of a function, this website gives some examples: 

http://www.coolmath.com/algebra/15-functions/06-finding-the-domain-01

p = perimiter

h = height

w = width

You have the inequality p<200, and the equation 2w+2h = p.

You can subsitute 2w+2h for p, so you get a new inequatlity of 2w+2h<200.

Since we know that w is 36, we can plug that value in: 2*36+2h<200  -->  72+2h<200.

Now we can solve this inequality for h: 

72+2h<200  -->  subtract 72 from both sides

2h<128        --> divide both sides by 2

h<64            --> Our answer is that the height of the frame must be less than 64 inches. 

−10 + 7𝑥 + 24 − 2  ---> combine like terms (add the numbers)

7x + 12 

If you have a set of numbers, the range is the difference between the largest number and the smallest number.

(Example - If your set was {3, 2, 7, 11, 12}, your range would be the largest number (12) - the smallest number (2).

[12-2 = 10, which is the range.]

1.05-z=1.65  --> subtract 1.05 from both sides

-z = 0.6 --> multiple both sides by -1

z = -0.6