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Black_prince [1.1K]
2 years ago
15

a square has a perimeter given by the expression 16x + 32y . Write an expression for the length of one side of the square

Mathematics
1 answer:
Lady_Fox [76]2 years ago
8 0
Perimeter=4*side so 

side=p/4 and we are told p=16x+32y so

s=(16x+32y)/4

s=4x+8y

s=4(x+2y)
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In 1875 an estimated twelve-and-a-half trillion locusts appeared in Nebraska.Write the number of locusts in scientific notation.
oksano4ka [1.4K]

Answer:

12.5 x 10^12

Step-by-step explanation:

12,500,000,000,000

12 zeroes following 12.5

3 0
3 years ago
Which of the following are considered measures of variability?
bixtya [17]

Answer:

the answers are <em>interquartile range</em> and <em>mean absolute deviation</em>

Step-by-step explanation:

6 0
3 years ago
How do I write a recursive formula for this sequence? 2, 6, 18, 54, 162, ... I already know the common difference is x3.
dmitriy555 [2]
For the sequence 2, 6, 18, 54, ..., the explicit formula is: an = a1 ! rn"1 = 2 ! 3n"1 , and the recursive formula is: a1 = 2, an+1 = an ! 3 . In each case, successively replacing n by 1, 2, 3, ... will yield the terms of the sequence. See the examples below.
4 0
3 years ago
What is the area of the trapezoid that is not drawn to scale??
netineya [11]
I think the answer would be 38in^2

Because the equation is : (a+b) *h / 2
A and b are both the bases of the trapezium

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(11+8)*4 / 2
(19)*4 / 2
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3 0
3 years ago
Read 2 more answers
Simplify each only using positive exponents:<br> 2x^-3 • 4x^2<br> 2x^4 • 4x^-3<br> 2x^3y^-3 • 2x
erica [24]
<h2>Answer:</h2>

\frac{2}{x}

\frac{x}{2}

\frac{4x^4}{y^3}

<h2>Step-by-step explanation:</h2>

a. 2x^-3 • 4x^2

To solve this using only positive exponents, follow these steps:

i. Rewrite the expression in a clearer form

2x⁻³ . 4x²

ii. The position of the term with negative exponent is changed from denominator to numerator or numerator to denominator depending on its initial position. If it is at the numerator, it is moved to the denominator. If otherwise it is at the denominator, it is moved to the numerator. When this is done, the negative exponent is changed to positive.

In our case, the first term has a negative exponent and it is at the numerator. We therefore move it to the denominator and change the negative exponent to  positive as follows;

\frac{1}{2x^3} . 4x^2

iii. We then solve the result as follows;

\frac{1}{2x^3} . 4x^2 = \frac{2}{x}

Therefore, 2x⁻³ . 4x² = \frac{2}{x}

b. 2x^4 • 4x^-3

i. Rewrite as follows;

2x⁴ . 4x⁻³

ii. The second term has a negative exponent, therefore swap its position and change the negative exponent to a positive one.

2x^4 . \frac{1}{4x^3}

iii. Now solve by cancelling out common terms in the numerator and denominator. So we have;

\frac{x}{2}

Therefore, 2x⁴ . 4x⁻³ = \frac{x}{2}

c. 2x^3y^-3 • 2x

i. Rewrite as follows;

2x³y⁻³ . 2x

ii. Change position of terms with negative exponents;

2x^3.\frac{1}{y^3} .2x

iii. Now solve;

\frac{4x^4}{y^3}

Therefore, 2x³y⁻³ . 2x = \frac{4x^4}{y^3}

8 0
3 years ago
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