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3 years ago

What does a-2+3=-2 equal

2 answers:

6 0

$Simplifying~
a + -2 + 3 = -2

Reorder the terms:
-2 + 3 + a = -2

Combine like terms: -2 + 3 = 1
1 + a = -2

Solving
1 + a = -2

Solving for variable 'a'.

Move all terms containing a to the left, all other terms to the right.

Add '-1' to each side of the equation.
1 + -1 + a = -2 + -1

Combine like terms: 1 + -1 = 0
0 + a = -2 + -1
a = -2 + -1

Combine like terms: -2 + -1 = -3
a = -3

Simplifying
a = -3$

Oliga [24]3 years ago

4 0

A=-3

a-2+3=-2
-3 -3

a-2=-5
+2 +2

a=-3

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2 years ago

Help me please I have no clue

Lynna [10]

Answer:

Step-by-step explanation:

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Here you’ll need to use the 30,60,90 degree triangle rule:

The opposite side of 30 degree will be y

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8 0

3 years ago

- 2/3x + 5(3/5 - 3/5x) - 12 divided 6 *2

Ira Lisetskai [31]

Answer: -3 2/3x - 1

Step-by-step explanation:

-2/3x + 5(3/5 - 3/5x) - 12 / 6 * 2

First distribute

-2/3x + 3 - 3x - 12 / 6 * 2

Now divide 12 by 6

-2/3x + 3 - 3x -2 * 2

Now multiply -2 by 2

-2/3x + 3 - 3x - 4

Now subtract 3x from -2/3x

-3 2/3x + 3 - 4

Now subtract 4 from 3

-3 2/3x - 1

4 0

2 years ago

Let C(n, k) = the number of k-membered subsets of an n-membered set. Find (a) C(6, k) for k = 0,1,2,...,6 (b) C(7, k) for k = 0,

vladimir1956 [14]

Answer:

(a) $C(6,0) = 1$ , $C(6,1) = 6$ , $C(6,2) = 15$ , $C(6,3) = 20$ , $C(6,4) = 15$ , $C(6,5) = 6$ , $C(6,6) = 1$ .

(b) $C(7,0) = 1$ , $C(7,1) = 7$ , $C(7,2) = 21$ , $C(7,3) = 35$ , $C(7,4) = 35$ , $C(7,5) = 21$ , $C(7,6) = 7$ , $C(7,7)=1$ .

Step-by-step explanation:

In this exercise we only need to recall the formula for C(n,k):

$C(n,k) = \frac{n!}{k!(n-k)!}$

where the symbol $n!$ is the factorial and means

$n! = 1\cdot 2\cdot 3\cdot 4\cdtos (n-1)\cdot n$ .

By convention 0!=1. The most important property of the factorial is $n!=(n-1)!\cdot n$ , for example 3!=1*2*3=6.

(a) The explanations to the solutions is just the calculations.

$C(6,0) = \frac{6!}{0!(6-0)!} = \frac{6!}{6!} = 1$

$C(6,1) = \frac{6!}{1!(6-1)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

$C(6,2) = \frac{6!}{2!(6-2)!} = \frac{6!}{2\cdot 4!} = \frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

$C(6,3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3!\cdot 3!} = \frac{5!\cdot 6}{6\cdot 6} = \frac{5!}{6} = \frac{120}{6} = 20$

$C(6,4) = \frac{6!}{4!(6-4)!} = \frac{6!}{4!\cdot 2!} = frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

$C(6,5) = \frac{6!}{5!(6-5)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

$C(6,6) = \frac{6!}{6!(6-6)!} = \frac{6!}{6!} = 1$ .

(b) The explanations to the solutions is just the calculations.

$C(7,0) = \frac{7!}{0!(7-0)!} = \frac{7!}{7!} = 1$

$C(7,1) = \frac{7!}{1!(7-1)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7$

$C(7,2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2\cdot 5!} = \frac{6!\cdot 7}{2\cdot 5!} = \frac{5!\cdot 6\cdot 7}{2\cdot 5!} = \frac{6\cdot 7}{2} = 21$

$C(7,3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!\cdot 4!} = \frac{6!\cdot 7}{6\cdot 4!} = \frac{5!\cdot 6\cdot 7}{6\cdot 4!} = \frac{120\cdot 7}{24} = 35$