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2 years ago

Write two expressions that are equivalent to the expression listed below. (X^1/3)^2

1 answer:

vladimir2022 [97]2 years ago

3 0

Answer

1) X⁽²/³⁾
2) (∛X)²

Explanation
The rule of indies states that..

A⁽ᵇ⁾∧ˣ ⁿ = A⁽ᵇˣⁿ⁾
X⁽₁/₃⁾∧₂ = X⁽²/³⁾

We can also write the same expressions as;

A¹/ⁿ = ⁿ√A

So,
X⁽₁/₃⁾∧₂ = (∛X)²

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Malik randomly picked two numbers from 1 to 9 (Includin 1 and 9). the same number could be picked more than once. The first of t

Tju [1.3M]

Answer:

2/9 = 0.22

Step-by-step explanation:

There are two ways to pick the first number odd and less than 5: 1 and 3.

With each of these, the second number drawn can be 1, 2, 3, 4, 5, 6, 7, 8 or 9.

This makes 18 total possibilities.

Out of these, the only ways to have a sum less than 5 are 1 and 1, 1 and 2, 1 and 3, and 3 and 1. This is 4 ways out of 18:

4/18 = 2/9 = 0.22

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2 years ago

Enya walked 2 km 309 m from school to the store. Then, she walked from the store to her home. If she

serg [7]

Her home was 2 km 691 m from the store. (5 km = 5000m --> 5000-2309)

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2 years ago

The average of your first 6 tests is 82. If your test average after your seventh test is an 80.5, what did you score on the seve

mart [117]

So average is of the first 6 tests is 82 well that means

(test1 + test2+ test3+ test4 + test5+test6)/6 = 82

so now let do some cross multiplliying

test1 + test2 + test3 + test 4 + test5 +test6 = 82*6

test1 + test2 + test3 + test 4 + test5 +test6 = 492

now lets see if we can find that 7th test score

(test1 + test2 + test3 + test4 + test5 +test6 + test7)/7 = 80.5

So look we found test1 + test2 + test3 + test4 + test5 +test6 to be 492 so lets substitute.

(492 + test7 )/7 = 80.5

test7 = (80.5*7)-492 = 71.5

7 0

2 years ago

a rectangular field has side lengths that measure 9/10 mile and 1/2 mile. What is the area of the field

GrogVix [38]

Answer:

9/20, or 0.45 mile.

Step-by-step explanation: $\\$

To find the area of a rectangle, multiply length times width. In this problem we are given both the length and width.

$\frac{9}{10}$ × $\frac{1\\}{2}$ is just 9/20, which is 0.45.

7 0

2 years ago

Find the distance from (4, −7, 6) to each of the following.

LenKa [72]

Answer:

(a) 6 units

(b) 4 units

(d) 9.22 units

(e) 7.21 units

(f) 8.06 units

Step-by-step explanation:

The distance d from one point (x₁, y₁, z₁) to another point (x₂, y₂, z₂) is given by;

d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

Now from the question;

(a) The distance from (4, -7, 6) to the xy-plane

The xy-plane is the point where z is 0. i.e

xy-plane = (4, -7, 0).

Therefore, the distance d is from (4, -7, 6) to (4, -7, 0)

d = √[(4 - 4)² + (-7 - (-7))² + (0 - 6)²]

d = √[(0)² + (0)² + (-6)²]

d = √(-6)²

d = √36

d = 6

Hence, the distance to the xy plane is 6 units

(b) The distance from (4, -7, 6) to the yz-plane

The yz-plane is the point where x is 0. i.e

yz-plane = (0, -7, 6).

Therefore, the distance d is from (4, -7, 6) to (0, -7, 6)

d = √[(4 - 0)² + (-7 - (-7))² + (6 - 6)²]

d = √[(4)² + (0)² + (0)²]

d = √(4)²

d = √16

d = 4

Hence, the distance to the yz plane is 4 units

(c) The distance from (4, -7, 6) to the xz-plane

The xz-plane is the point where y is 0. i.e

xz-plane = (4, 0, 6).

Therefore, the distance d is from (4, -7, 6) to (4, 0, 6)

d = √[(4 - 4)² + (-7 - 0)² + (6 - 6)²]

d = √[(0)² + (-7)² + (0)²]

d = √[(-7)²]

d = √49

d = 7

Hence, the distance to the xz plane is 7 units

(d) The distance from (4, -7, 6) to the x axis

The x axis is the point where y and z are 0. i.e

x-axis = (4, 0, 0).

Therefore, the distance d is from (4, -7, 6) to (4, 0, 0)

d = √[(4 - 4)² + (-7 - 0)² + (6 - 0)²]

d = √[(0)² + (-7)² + (6)²]

d = √[(-7)² + (6)²]

d = √[(49 + 36)]

d = √(85)

d = 9.22

Hence, the distance to the x axis is 9.22 units

(e) The distance from (4, -7, 6) to the y axis

The x axis is the point where x and z are 0. i.e

y-axis = (0, -7, 0).

Therefore, the distance d is from (4, -7, 6) to (0, -7, 0)

d = √[(4 - 0)² + (-7 - (-7))² + (6 - 0)²]

d = √[(4)² + (0)² + (6)²]

d = √[(4)² + (6)²]

d = √[(16 + 36)]

d = √(52)

d = 7.22

Hence, the distance to the y axis is 7.21 units

(f) The distance from (4, -7, 6) to the z axis

The z axis is the point where x and y are 0. i.e

z-axis = (0, 0, 6).

Therefore, the distance d is from (4, -7, 6) to (0, 0 6)

d = √[(4 - 0)² + (-7 - (0))² + (6 - 6)²]

d = √[(4)² + (-7)² + (0)²]

d = √[(4)² + (-7)²]

d = √[(16 + 49)]

d = √(65)

d = 8.06

Hence, the distance to the z axis is 8.06 units

5 0

2 years ago