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erastovalidia [21]
3 years ago
11

What is the difference? StartFraction x Over x squared minus 16 EndFraction minus StartFraction 3 Over x minus 4 EndFraction Sta

rtFraction 2 (x + 6) Over (x + 4) (x minus 4) EndFraction StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction StartFraction x minus 3 Over (x + 5) (x minus 4) EndFraction StartFraction negative 2 (x minus 6) Over (x + 4) (x minus 4) EndFraction
Mathematics
2 answers:
blagie [28]3 years ago
8 0

Answer:

Cara multiplied only 4 and 7 together and did not multiply 4 and 13.

Step-by-step explanation:

katovenus [111]3 years ago
5 0

Answer:

The option "StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction" is correct

That is \frac{-2(x+6)}{(x+4)(x-4)}

Therefore \frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}

Step-by-step explanation:

Given problem is StartFraction x Over x squared minus 16 EndFraction minus StartFraction 3 Over x minus 4 EndFraction

It can be written as below :

\frac{x}{x^2-16}-\frac{3}{x-4}

To solve the given expression

\frac{x}{x^2-16}-\frac{3}{x-4}

=\frac{x}{x^2-4^2}-\frac{3}{x-4}

=\frac{x}{(x+4)(x-4)}-\frac{3}{x-4}  ( using the property a^2-b^2=(a+b)(a-b) )

=\frac{x-3(x+4)}{(x+4)(x-4)}

=\frac{x-3x-12}{(x+4)(x-4)} ( by using distributive property )

=\frac{-2x-12}{(x+4)(x-4)}

=\frac{-2(x+6)}{(x+4)(x-4)}

\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}

Therefore \frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}

Therefore the option "StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction" is correct

That is \frac{-2(x+6)}{(x+4)(x-4)}

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andrew11 [14]

Answer:

8 = b - 1 + 8b

group like terms

8 + 1 = b + 8b

9 = 9b cus b standing alone me 1b

divide both side by 9

<u>9</u><u> </u> = <u>9</u><u>b</u>

9 9

b = 1

7 0
2 years ago
Question 3 (Multiples and Factors] Three numbers are given below. Use prime factorisation to determine the HCF and LCM 1848 132
Ilia_Sergeevich [38]

Prime factorization involves rewriting numbers as products

The HCF and the LCM of 1848, 132 and 462​ are 66 and 1848 respectively

<h3>How to determine the HCF</h3>

The numbers are given as: 1848, 132 and 462

Using prime factorization, the numbers can be rewritten as:

1848 = 2^3 * 3 * 7 * 11

132 =  2^2 * 3 * 11

462 = 2 * 3 * 7 * 11

The HCF is the product of the highest factors

So, the HCF is:

HCF = 2 * 3 * 11

HCF = 66

<h3>How to determine the LCM</h3>

In (a), we have:

1848 = 2^3 * 3 * 7 * 11

132 =  2^2 * 3 * 11

462 = 2 * 3 * 7 * 11

So, the LCM is:

LCM = 2^3 * 3 * 7 * 11

LCM  = 1848

Hence, the HCF and the LCM of 1848, 132 and 462​ are 66 and 1848 respectively

Read more about prime factorization at:

brainly.com/question/9523814

4 0
2 years ago
The model below represents the equation shown. What is the value of x?
Angelina_Jolie [31]

Answer:

Step-by-step explanation:

4 0
3 years ago
Factories 54m3n + 81m4n2 b) 15x2y3z + 25x3y2z + 35x2y2z​
Oliga [24]

<u>Part a)</u>

Given the expression

54m^3n\:+\:81m^4n^2

Apply exponent rule:    a^{b+c}=a^ba^c

54m^3n\:+\:81m^4n^2=54m^3n+81m^3mnn     ∵ m^4n^2=m^3mnn

Rewrite 81 as 3 · 27

Rewrite 54 as 2 · 27

                             =2\cdot \:27m^3n+3\cdot \:27m^3mnn

Factor out the common term:   27m³n

                              =27m^3n\left(2+3mn\right)      

Therefore,

54m^3n\:+\:81m^4n^2=54m^3n+81m^3mnn=27m^3n\left(2+3mn\right)

<u>Part B)</u>

Given the expression

15x^2y^3z+25x^3y^2z+35x^2y^2z

Apply exponent rule:    a^{b+c}=a^ba^c

15x^2y^3z+25x^3y^2z+35x^2y^2z=15x^2y^2yz+25x^2xy^2z+35x^2y^2z

Rewrite as

                                               =3\cdot \:5y^2x^2zy+5\cdot \:5y^2x^2zx+7\cdot \:5y^2x^2z

Factor out common term 5y²x²z

                                                =5y^2x^2z\left(3y+5x+7\right)

Therefore,

15x^2y^3z+25x^3y^2z+35x^2y^2z=5y^2x^2z\left(3y+5x+7\right)

8 0
3 years ago
In the supermarket a loaf of bread costs 37p how many loaves can david buy with a 2 coin ? How much money will be left over?
Rasek [7]

Answer: He can buy 5 loaves of bread.

After buying 5 loaves 15 p will be left.

Step-by-step explanation:

Given, In the supermarket a loaf of bread costs 37p .

To find: How many loaves can David buy with a \pounds 2 coin?

Since \pounds 1=100\text{ pence}

Then, \pounds 2=200\text{ pence}

Number of loaves he can buy = (Amount he has) ÷ (Cost of a loaf of bread)

= 200 p ÷ 37 p

=5\dfrac{15}{37}

i.e. he can buy 5 loaves of bread and 15 p will be left.

Hence, He can buy 5 loaves of bread.

After buying 5 loaves 15 p will be left.

6 0
3 years ago
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