The value of the pronumeral

Find the difference between the result of one sixth of product of 10 and 3 multiplied by 4 and 30

kherson [118]

Answer:

10 or -10,

Read the explanation

Step-by-step explanation:

Rewrite this problem as a numerical expression. As per the wording of this problem, there can be two expressions derived.

1. $((\frac{1}{6}(10*3)*4)-30$

2. $30-((\frac{1}{6}(10*3)*4)$

Simplify, remember the order of operations. The order of operations is the sequence by which one is supposed to perform operations in a numerical expression. This order is the following:

1. Parenthesis

2. Exponents

3. Multiplication or division

4. Addition or Subtraction

Use this sequence when simplifying and solving the expression:

Expression 1

$((\frac{1}{6}(10*3)*4)-30\\\\=((\frac{1}{6}(30)*4)-30\\\\=(5*4)-30\\\\=20 - 30\\\\= -10$

Expression 2

$30-((\frac{1}{6}(10*3)*4)\\\\=30-((\frac{1}{6}(30)*4)\\\\=30-(5*4)\\\\= 30-20\\\\= 10$

6 0

3 years ago

Two rectangles are the same one has a length of 11 cm and a width of 10 cm and the other has a width of 7 cm find the length of

ser-zykov [4K]

Answer:

D) 7.7 cm

Step-by-step explanation:

L/W=11cm/10cm=x/7cm

cross multiply to find x

10x=77

divide 10 by both sides

x=7.7

8 0

4 years ago

Why is 4 and 44 not proportional

Oksanka [162]

Because 44 is a bigger number than 4 it is 11 times bigger so the difference is larger than a proportional number

5 0

3 years ago

group of friends wants to go to the amusement park. They have $366.50 to spend parking and admission. Parking is $19, and ticket

Lostsunrise [7]

Step 1:
Subtract $19 from total
366.50 - 19 = 347.5

Step 2:
Divide this total by the ticket price
347.50 / 34.75 = 10

Step 3:
You have your answer
10 people can go to the amusement park

5 0

3 years ago

(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

3 years ago

The value of the pronumeral​

The value of the pronumeral