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4 years ago

(x^-2/y^-3)^2 / (y/x^2)^-3

2 answers:

6 0

$\frac{ (\frac{ x^{-2} }{ y^{-3} })^2 }{ (\frac{y}{ x^{2} })^{-3} } = \frac{ (\frac{ x^{-4} }{ y^{-6} }) }{ (\frac{ y^{-3} }{ x^{-6} }) }= \frac{ x^{-4}* x^{-6} }{ y^{-3}* y^{-6} } = \frac{ x^{-10} }{ y^{-9} }$

azamat4 years ago

3 0

(x-²/ y-³)² = (x-²/y-³)(x-²/y-³) = (y/ x²)(y/ x²) = (y²/ x^4) = y² · x^4 = 1
(y/ x²)-³ (y/x²)(y/x²) (x²/y³)(x²/y³) (x^4/ y^6) x^4 · y^6 y^4

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the cost of a phone call is c pence which is calculated using the formula c=20+5m where m is the number of minutes the phone cal

OLga [1]

I believe the answer is 25 because it says that 2)m=5 or it may be 30 because 5times 2 equals 10 and 10plus 20 gives you 30

4 0

4 years ago

Which conjecture is possible to prove is congruent? A. All quadrilaterals with 4 equal angles B. All triangles with 3 equal angl

igor_vitrenko [27]

Answer: D. All triangles with 3 equal sides

Step-by-step explanation:

Remember, congruence means that the sides are equal and also the corresponding angles:

A) Quadrilaterals with 4 equal angles.

Here we can have, for example, a lot of different rectangles and squares, because all of them have only 45° angles, but a square is not congruent with a rectangle, so this option is not correct.

B) Triangles with 3 equal angles.

Think in the triangles, the 3 angles are 60°,

But we can craft this angle with a lot of different side lengths, so this is not the correct option, so not all the triangles with equal angles are congruent, then this option is incorrect.

C) Quadrilaterals with 4 equal sides: Here we can have a square, or a rhombus, or a parallelogram, so this option is also incorrect.

D) Triangles with 3 equal sides:

Now, we have the set of all the triangles where the length of the 3 sides is the same.

Now, if you have 3 knowed lenghts (the ones shared by the set of triangles) you can only create one triangle with them, this is known as the SSS criterion of congruence for triangles. This is the correct option.

7 0

3 years ago

The balance of errands average balance checking account at the beginning of last cycle was $400 and the only transaction for the

Tom [10]

Answer:

$300

Step-by-step explanation:

Here we are given:

Balance at the beginning of last cycle(first half) = $400

Balance halfway through the cycle(second half cycle) = $200

To find the amount Aaron's checking account pay interest last cycle, let's find the average balance.

Use the formula for average:

$\frac{Bal 1 + Bal 2}{2}$

$\frac{400 + 200}{2}$

$\frac{600}{2} = 300$

Average = $300

Therefore, the amount Aaron's checking account pay interest last cycle is $300

8 0

3 years ago

1. Which of the following is equivalent to x5y2/xy2 when x ≠ 0 and y ≠ 0?

Sedaia [141]

Q1. The answer is D. x4

Let's first rewrite the expression:
x⁵y²/xy² = x⁵/x * y²/y²

Using the rule xᵃ/xᵇ = x(ᵃ⁻ᵇ), we can write the expression as following:
x⁵y²/xy² = x⁵/x * y²/y² = x⁵⁻¹ * y²⁻² = x⁴ * y⁰ = x⁴ * 1 = x⁴

Thus, the correct answer is D.

Q2. The answer is A. 5(5^3/2/5)^2

125 in the form of exponent is 5³.
125 = 5³
Now, let's calculate all choices.

The rules we will use are:
xᵃ * xᵇ = x(ᵃ⁺ᵇ)
xᵃ/xᵇ = x(ᵃ⁻ᵇ)
(xᵃ)ᵇ = xᵃ*ᵇ

A. 5(5³/2/5)² = 5 * (5³ * 5/2)²
= 5 * (5³⁺¹/2)²
= 5 * (5⁴/2)²
= 5 * (5⁴)²/(2)²
= 5 * 5⁴*²/4
= 5 * 5⁸ / 4
= 5¹⁺⁸ / 4
= 5⁹/4
≠ 5³ ≠ 125

B. (5³/5⁴)⁻³ = (5³⁻⁴)⁻³
= (5⁻¹)⁻³
= 5⁽⁻¹⁾*⁽⁻³⁾
= 5³
= 125

C. 5⁻²/5⁻⁵ = 5⁽⁻²⁾⁻⁽⁻⁵)
= 5⁽⁻²⁾⁺⁵
= 5³
= 125

D. 5(5⁵/5³) = 5 * 5⁵⁻³
= 5 * 5²
= 5¹⁺²
= 5³
= 125

Therefore, the only expression that is not equal to 125 is A.

Q3. The answer is 63x5

Let's check all choices
The rules we will use are:
xᵃ * xᵇ = x(ᵃ⁺ᵇ)
xᵃ/xᵇ = x(ᵃ⁻ᵇ)
(xᵃ)ᵇ = xᵃ*ᵇ

A. 6³x
6³x/6x⁵ = 6³/6 * x/x⁵
= 6³⁻¹ * x¹⁻⁵
= 6²x⁻⁴
= 36x⁻⁴
≠ 36

B. 6³x⁵
6³x⁵/6x⁵ = 6³/6 * x⁵/x⁵
= 6³⁻¹ * x⁵⁻⁵
= 6² * x⁰
 = 36 * 1
= 36

C. 6x⁵
6x⁵/6x⁵ = 1
≠ 36

D. 6⁷x⁵
6⁷x⁵/6x⁵ = 6⁷/6 * x⁵/x⁵
= 6⁷⁻¹ * x⁵⁻⁵
= 6⁶ * x⁰
= 46656 * 1
≠ 36

Therefore, the correct choice is B.

Q4. The answer is

We will use the rule: xᵃ/xᵇ = x(ᵃ⁻ᵇ)

5.4 x 10¹²/1.2 x 10³ = 5.4 / 1.2 x 10¹²/10³
= 4.5 x 10¹²⁻³
= 4.5 x 10⁹

Q5. The answer is B. To subtract powers with the same base, divide the exponents

Some of the rules regarding operations with exponents are:
xᵃ/xᵇ = x(ᵃ⁻ᵇ) - choice A
xᵃ * xᵇ = x(ᵃ⁺ᵇ) - choice C
(xᵃ)ᵇ = xᵃ*ᵇ - choice D

Through the process of elimination, choice B is not true.

8 0

3 years ago

Read 2 more answers

(y^11)(y^2)^8 + y^x what is the value of x?

Harman [31]

I think it may be 3. not sure though

3 0

3 years ago