3(4-2(5-3)+10+5×1) = n please show the solution?

3(4-2(5-3)+10+5×1) = n

Solution:

3(4-2(2)+10+5×1) = n

3(4-4+10+5×1) = n

3(4-4+10+5) = n

3(0+10+5) = n

3(10+5) = n

3(15) = n

45 = n

Remember: BEDMAS is the order for solving these types of problems.

Brackets first

Exponents next

Division or

Multiplication (do division/multiplication in the order they appear)

Addition or

Subtraction (do addition/subtraction in the order they appear)

Order of operations. First the inside parenthesis (5-3), then the first multiplication inside the parenthesis (2×2), then the other (5×1), then the adding and subtracting from left to right, finally multiply the result by 3.

3(4-2(5-3)+10+5×1)=n

3(4-2(2)+10+5×1)=n

3(4-4+10+5×1)=n

3(4-4+10+5)=n

3(0+10+5)=n

3(15)=n

45=n

a² + b² = c² a + b + c = ab Multiply the two factors by 2 and upload a² + b² on the two factors. a² + b² + 2a + 2b + 2c = a² + 2ab + b² a² + b² + 2a + 2b + 2c = (a + b)² a² + b² + 2c = (a + b)(a + b – 2) c² + 2c = (a + b)(a + b – 2) c(c + 2) = (a + b)(a + b – 2) bear in mind that a + b > c for any triangle. hence c + 2 > a + b – 2. c + 4 > a + b > c c² + 8c + sixteen > a² + b² + 2ab > c² 4c + 8 > ab > 0 8 > ab – 4?(a² + b²) > -4?(a² + b²) With this constraint, we can wager all the possible values and notice that are the only ones that paintings. it may be a = 3, b = 4 or a = 4, b = 3. of direction, there is particularly some guessing in contact right here. however the terrific element right it incredibly is that ?(a² + b²) is an integer. we will additionally ought to apply value of replace as area of the failings we are going to use for the guessing. —– EDIT: d = a + b From our final equation: (c + d) / (d – 2) = (c + d) / c (c + d)(a million / (d – 2) – a million / c) = 0 a million / (d – 2) = a million/c c = a + b – 2 the only acceptable triangle with this high quality is (3, 4, 5).

=3(4-2(5-3)+10+5×1)

use the PEMDAS method, which represents the order of operations in such complex problems:

P: parentheses. simplify the innermost stuff first.

=3(4-2(2)+10+5×1)

E: exponents. however, none are shown here, so we go to the next operation…

M: multiplication. but keep in mind that, in this case, you still have to simplify those stuff inside the parentheses…

=3(4-4+10+5×1)

=3(4-4+10+5)

D: division. not applicable in this problem…

A: addition. again, we’re still not finished doing the stuff inside the parentheses…

=3(4+6+5)

=3(15)

S: subtraction, but we inadvertedly dealt with that earlier, as you can see.

having finished the inner parts, only two terms and one operation are left. do that, you get 45.

If i can remember all the way back to high school…i believe this is the correct way to answer this question.

3(4-2(5-3)+10+5)

3(4-2(2)+10+5)

3(4-4+10+5)

3(0+10+5)

3(15)

45

Remember the rule BIMDAS

= 3(4-2(2)+10+5×1)=n

= 3(2(2)+10+5)=n

= 3(4+15)=n

= 3(19)=n

n=57

Ok, I know that’s not the correct answer…

(5-3)=2

(4-2×2+10+5X1)

(4-4+10+5) =15

3×15

45

n=45

Using order of precedence of operators and parens here is what I get:

3(4-2(5-3)+10+5)

3(4-2(2)+10+5)

3(4-4+10+5)

3(0+10+5)

3(15)

45

45 = n

3(4-2(2)+10+5)=n

3(4-4+10+5)=n

3(15)=n

45=n

This is done through order of operations aka PEMDAS (parentheses, exponents, multiplication, division, addition, subtraction)

hope this helps 🙂

solve 1st the inside3( 4-2)(5-3)+50)=3{(2)(2)+50}=54×3=162