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Sonbull [250]
3 years ago
5

Why does the multiplication property of equality not allow us to divide both sides of an equation by zero?​

Mathematics
1 answer:
Anna71 [15]3 years ago
6 0

First of all, recall that division by zero is undefined; it's nonsensical; it's just not allowed.So zero certainly needs to be excluded when dividing.

But what about multiplying by zero?

The problem is that multiplying by zero can change the truth of an equation:

It can take a false equation to a true equation.

To see this, consider the false equation ‘

2 = 3

Multiplying both sides by zero results in the new equation

2 ⋅ 0 = 3 ⋅0 (that is, ‘0 = 0’), which is true.

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Find a symbolic representation for f ^-1​(x).
Alexus [3.1K]

y = (9x) ^ (1/3)

exchange x and y  then solve for y

x = (9y) ^ (1/3)

cube each side

x^3 = 9y

divide each side by 9

1/9 x^3 = y

the inverse function is 1/9 x^3

4 0
2 years ago
In △ABC, m∠A = α, m∠B = β, m∠C = γ. AB = c, AC = b, BC = a. Find the areas of the triangles if the following parts are given.
Mazyrski [523]
A = c/sin C, and b by the proportion b/sin B = c/sin C.

a/sin 37° = 4380/sin 109°
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6 0
3 years ago
FIRST ANSWER GETS BRAINLIEST
Anarel [89]

Answer:

4/15

Step-by-step explanation:

You would do it as follows

I-8/15I reduce the fraction by 2

I-4/25I the absolute fraction is always positive

So the solution is

4/25

Alternative forms are:

0.16 or (2/5)^2 (Just incase)

5 0
2 years ago
Read 2 more answers
Is 0.75 greater than 0.100?
Naya [18.7K]
Yes because you only need to look at the first decimal for problems like these and the first number you see a seven and a one and seven is greater than one
8 0
3 years ago
Triangle PQR has coordinates P(–8, 3), Q(–8, 6), and R(–3, 6). If the triangle is translated by using the rule (x, y) right-arro
omeli [17]

Answer:

d

Step-by-step explanation:

The rule (x, y ) → (x + 4, y - 6 )

means add 4 to the original x- coordinate and subtract 6 from the original y- coordinate, thus

P(- 8, 3 ) → P'(- 8 + 4, 3 - 6 ) → P'(- 4, - 3 )

Q(- 8, 6 ) → Q'(- 8 + 4, 6 - 6 ) → Q'(- 4, 0 )

R(- 3, 6 ) → R'(- 3 + 4, 6 - 6 ) → R'(1, 0 )

6 0
3 years ago
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