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allochka39001 [22]
4 years ago
9

Simplify X (x^2+6y^2)

Mathematics
1 answer:
andrew11 [14]4 years ago
5 0
Simplify by opening the bracket;
x(x^2+6y^2)
=x^{3} +6 y^{3}
You might be interested in
Can someone plzz help am stuck
navik [9.2K]
It is only asking for the 3/4 length buttons
There are 3 of them

Lining them up means adding them together

3/4 + 3/4 + 3/4 = 9/4, or 2 1/4

2 1/4 should be your answer

hope this helps
3 0
3 years ago
Read 2 more answers
09:57:09
Sveta_85 [38]
8 cm and 9 cm is the correct answer
8 0
3 years ago
Recall the scenario about Eric's weekly wages in the lesson practice section. Eric's boss have been very impressed with his work
Solnce55 [7]

Answer:  

1)\quad f(x)=\bigg\{\begin{array}{ll}12x&0\leq x

2) D: x = [0, 24]

3) R: y = [0, 384]

4) see graph

<u>Step-by-step explanation:</u>

Eric's regular wage is $12 per hour for all hours less than 9 hours.

The minimum number of hours Eric can work each day is 0.

f(x) = 12x    for   0 ≤ x < 9

Eric's overtime wage is $18 per hour for 9 hours and greater.

The maximum number of hours Eric can work each day is 24 (because there are only 24 hours in a day).

f(x) = 18(x - 8) + 12(8)

    = 18x - 144 + 96

    = 18x - 48           for 9 ≤ x ≤ 24

The daily wage where x represents the number of hours worked can be displayed in function format as follows:

f(x)=\bigg\{\begin{array}{ll}12x&0\leq x

2) Domain represents the x-values (number of hours Eric can work).

The minimum hours he can work in one day is 0 and the maximum he can work in one day is 24.

D:  0 ≤ x ≤ 24        →        D: x = [0, 24]

3) Range represents the y-values (wage Eric will earn).

Eric's wage depends on the number of hours he works. Use the Domain (given above) to find the wage.

The minimum hours he can work in one day is 0.

f(x) = 12x

f(0) = 12(0)

     =  0

The maximum hours he can work in one day is 24 <em>(although unlikely, it is theoretically possible).</em>

f(x) = 18x - 48

f(24) = 18(24) - 48

       = 432 - 48

       = 384

D:  0 ≤ y ≤ 384        →        D: x = [0, 384]

4) see graph.

Notice that there is an open dot at x = 9 for f(x) = 12x

and a closed dot at x = 9 for f(x) = 18x - 48

6 0
3 years ago
A stack of books is 21 inches tall. Each book is 1 3/4 inches. How many books are there?
Maru [420]
There are twelve books
3 0
3 years ago
Read 2 more answers
The ratio of the number of dogs the number of cats is 4:5. There are 270 animals on the rescue Farm. The number of dogs is 4/5.
mafiozo [28]

Answer:

<em>Here is the complete question:</em>

The ratio of the number of dogs to the number of cats is 4:5. There are 270 animals on the rescue farm. for each of the following statements, explain whether statements is true or false and why:

a. The number of dogs is 4/5 the number of cats.

b. 4/5 of the pets at the farm are dogs.

c. There are exactly 30 more cats than dogs.

d. There are exactly 30 dogs at the farm.

e. 5/9 of the pets at the farm are cats.

a,c and e are true statements.

Step-by-step explanation:

Let the number of dogs be "d' and number of cats be "c".

According to the question:

⇒ c+d=270   ...equation (i)

Arranging the ratio:

⇒ \frac{4}{5} = \frac{d}{c}

⇒ 4c=5d

⇒ c=(\frac{5}{4})d    ...equation (ii)

Plugging (ii) in equation (i).

⇒ c+d=270

⇒ \frac{5}{4} d+d=270

⇒ \frac{5d+4d}{4} =270

⇒ 9d=270\times 4

⇒ d=\frac{270\times 4}{9}

⇒ d=120

Number of dogs "d" = 120

Number of cats "c" = (270-120) = 150

Lets check each statement:

a.

The number of dogs is 4/5 the number of cats.

⇒ \frac{4}{5} \times 150

⇒ \frac{4\times 150}{5}

⇒ 120

Statement is true.

b.

4/5 of the pets at the farm are dogs.

⇒ \frac{4}{5} \times 270

⇒ 216

Number of dogs = 120

So the above statement is false.

c.

There are exactly 30 more cats than dogs.

⇒ c-d =30

⇒ 150-120=30

Statement is true.

d.

There are exactly 30 dogs at the farm.

False, as there are 120 dogs

e.

5/9 of the pets at the farm are cats.

⇒ \frac{5}{9} \times 270

⇒ 150

Statement is true.

So,

Statement  a,c and e are true as we could observe in the explanation.

8 0
3 years ago
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