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2 years ago

14

How can we isolate the variable c

1 answer:

Vlada [557]2 years ago

8 0

Well depending on the equation you look at both sides and eliminate everything from both using addition, subtraction, multiplication, and division.

Ex:

3c-4=7 first add the 4 because you use the opposite type of operation so you add it to 7 which gives you:

3c=11 next you have to divide to get it alone

3c/3=11/3

which gives you

c= 3.66666666667

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Question 12 (2 points)

elena55 [62]

Answer:

A box plot is drawn with end points at 24 and 49.The box extends from 28 to 44 and a vertical line is drawn inside the box at 34.

Step-by-step explanation:

Ordering the data given :

24,28,32,34,40,44,49

We can calculate the 5 number summary required to give the appropriate boxplot that can be produced :

Minimum = 24

Maximum = 49

Median = 1/2(n+1)th term

n = 7

Median = 1/2(8) = 4th term

Median = 34

Lower quartile, Q1 = 1/4(n+1)th term

n = 7

1/4(8) = 2nd term

Q1 = 28

Upper quartile : 3/4(n+1)th term

n = 7

Q3 = 3/4(8) = 6th term

Q3= 44

3 0

2 years ago

Find an upper limit for the zeroes 2x^4 -7x^3 + 4x^2 + 7x - 6 = 0

erik [133]

<u>Answer-</u>

2 is the upper limit for the zeros.

<u>Solution-</u>

The given function f(x) is,

$2x^4 -7x^3 + 4x^2 + 7x - 6 = 0$

For calculating the zeros,

$\Rightarrow f(x)=0$

$\Rightarrow 2x^4 -7x^3 + 4x^2 + 7x - 6 = 0$

$\Rightarrow 2x^4-4x^3-3x^3+6x^2-2x^2+ 4x+3x-6=0$

$\Rightarrow 2x^3(x-2)-3x^2(x-2)-2x(x-2)+3(x-2)=0$

$\Rightarrow (x-2)(2x^3-3x^2-2x+3)=0$

$\Rightarrow (x-2)(x^2(2x-3)-1(2x-3))=0$

$\Rightarrow (x-2)(x^2-1)(2x-3)=0$

$\Rightarrow (x-2)(x+1)(x-1)(2x-3)=0$

$\Rightarrow x-2=0,\ x+1=0,\ x-1=0,\ 2x-3=0$

$\Rightarrow x=2,\ x=-1,\ x=1,\ x=\frac{3}{2}$

From all the 4 roots, it can be obtained that 2 is the greatest zero.

7 0

2 years ago

2. (- 6x + 3) + (x - 7)

liq [111]

Answer:

-5x-4

Step-by-step explanation:

You first have to combine like terms. (-6x + x) + (+3 -7) = -5x-4

5 0

2 years ago

Read 2 more answers

Whats the roots of the following three problems:

klemol [59]

QUESTION 1

The given function is

$f(x)=\frac{(x-3)(x-1)}{(x-1)(x+2)}$

We simplify to get;

$f(x)=\frac{x-3}{x+2}$

This function is equal to zero when $x-3=0$

$\Rightarrow x=3$

QUESTION 2

The given function is

$f(x)=\frac{5x^2-10x+5}{2x^2-5x+3}$

$f(x)=\frac{5(x^2-2x+1)}{2x^2-5x+3}$

We factor to get;

$f(x)=\frac{5(x-1)^2}{(2x-3)(x-1)}$

$f(x)=\frac{5(x-1)}{(2x-3)}$

This function equals zero when

$5(x-1)=0$

$x=1$

QUESTION 3

The given function is

$f(x)=\frac{x^3-8}{x^2-6x+8}$

We factor to get,

$f(x)=\frac{(x-2)(x^2+2x+4)}{(x-2)(x+4)}$

$f(x)=\frac{x^2+2x+4}{x+4}$

The function equals zero when $x^2+2x+4=0$

$D=b^2-4ac$

$D=2^2-4(1)(4)$

$D=-12$

Hence the equation has no real roots.

The complex roots are

$x=-\sqrt{3}i-1$ or $x=-1+\sqrt{3}i$

4 0

2 years ago

-2X-3-4Х<br><br> A) -6x -3<br> B) -3 -2x -4x<br> C) -x + 4x

Pavlova-9 [17]

The answer is A, make sure to combine like terms

5 0

2 years ago