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3 years ago

"If a = − 9 and b = − 6, show that (a−b) ≠ (b−a)."

Mathematics

1 answer:

denis-greek [22]3 years ago

7 0

Answer:

Step-by-step explanation:

LHS a - b = -9 - (-6) = -9 +6 = -3

RHS b-a = -6 - (- 9) = -6 +9 = 3

as LHS not equal to RHS

a-b not equal to b-a

Thus proven

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(no links) PLS HELP AGAIN I LOVE U

Minchanka [31]

y = 3(x + 4)^2 + 31

Step-by-step explanation:

We can convert the given quadratic equation into its vertex form by completing the square:

y = 3x^2 + 24x + 43

= 3(x^2 + 8x) + 43

= 3(x^2 + 8x + 4) + 31

= 3(x + 4)^2 + 31

This is the vertex form of the given quadratic equation with (-4, 31) as its vertex

4 0

3 years ago

What percent of 50 is 9?

d1i1m1o1n [39]

9 is 18% of 50. So, 50 percent also written as 50% means 50 out of 100 parts. Since there are exactly two 50s in 100, fifty percent is the same number as one half. 50 percent=0.5 in but as a fraction, its 1/2.

8 0

3 years ago

this is the rust of the questions please please some one helps me because I really try my best to solve it . thank you

lyudmila [28]

Answer:

x = 24.

r $\ne$ 0.

Step-by-step explanation:

2. The given equation is:

$\frac{1}{2} (x - 4) = \frac{1}{3} x + 2$

a) To eliminate the fractions multiply the equation throughout by the LCM of the denominators of the fraction. In this case, the LCM of (2, 3). The LCM is 6. So, multiply the entire equation by 6.

b) Half of the difference between an integer and 4 equals the sum of one - third of the integer and 2. Find the integer.

c) We have the equation:

$\frac{1}{2} (x - 4) = \frac{1}{3} x + 2$

Multiplying throughout by 6, we get:

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

$\implies 3(x - 4) = 2x + 12$

$\implies 3x - 12 = 2x + 12$

$\implies x = 24$

Therefore, the solution of the equation is 24.

3. The given equation is: $ry + s = tx - m$

To solve for y:

We can rearrange the equation as:

$ry = tx - m - s$

$\implies y = \frac{tx - m - s}{r}$

or, $y = \frac{tx - (m + s)}{r}$

Note that we have to impose a condition on variable $r$ . It would be that $r$ can never be zero. i.e., $r \ne 0$ . Otherwise, the value of $y$ would be undefined.