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4 years ago

A person who weighs 100 pounds on Earth weighs 16.6 lb on the moon. Which variable is the independent variable?

Mathematics

2 answers:

scoundrel [369]4 years ago

8 0

The person who weighs 100 pounds on earth.

Because in order to find the pounds on the moon you have to know how much they weigh on Earth which makes the pounds on the moon the dependent variable and the pounds on earth the dependent.

AleksandrR [38]4 years ago

7 0

The person who weighs 100 pounds

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If the product of the extremes is 8,then the geometric mean is. A) 2. B) 2√2. C) 4

Rainbow [258]

if,

p/x=x/q

then cross-multiplying gives x^2=pq

x^2=pq. Taking the square root of both sides, we get x=√pq as the geometric mean of p and q.

Hence in the given question the product of p and q is 8.

so,

x=√8

x=2√2

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3 years ago

A home gardener estimates that 18 apple trees will produce an average yield of 80 apples per tree. But because of the size of th

aliina [53]

19 trees should be planted to maximize the total

<h3>How many trees should be planted to maximize the total</h3>

From the question, we have the following parameters:

Number of apples, x = 18

Yield, f(x) = 80 per tree

When the number of apple trees is increased (say by x).

We have:

Trees = 18 + x

The yield decreases by four apples per tree.

So, we have

Yield = 80 - 4x

So, the profit function is

P(x) = Apples * Yield

This gives

P(x) = (18 + x) *(80 - 4x)

Expand the bracket

P(x) = 1440 - 72x + 80x - 4x^2

Differentiate the function

P'(x) = 0 - 72 + 80 - 8x

Evaluate the like terms

P'(x) = 8 - 8x

Set P'(x) to 0

8 - 8x = 0

Divide through by 8

1 - x = 0

Solve for x

x = 1

Recall that:

Trees = 18 + x

So, we have

Trees = 18 + 1

Evaluate

Trees = 19

Hence, 19 trees should be planted to maximize the total

Read more about quadratic functions at:

brainly.com/question/12120831

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1 year ago

2 If A= 4 3 find A1 using elementary row operations.

MA_775_DIABLO [31]

Answer: $A^{-1}=\left[\begin{array}{cc}\frac{3}{2}&-\frac{1}{2}\\-2&1\end{array}\right]$

Step-by-step explanation:

$\left[\begin{array}{cc}2&1\\4&3\end{array}\right]=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

$\dfrac{1}{2}Row\ 1\rightarrow\left[\begin{array}{cc}1&\frac{1}{2}\\4&3\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}&0\\0&1\end{array}\right]$

$Row\ 2 -4 \ Row\ 1\rightarrow \left[\begin{array}{cc}1&\frac{1}{2}\\0&1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}&0\\-2&1\end{array}\right]$

$Row\ 1-\dfrac{1}{2}\ Row\ 2 \rightarrow \left[\begin{array}{cc}1&0\\0&1\end{array}\right]=\left[\begin{array}{cc}\frac{3}{2}&-\frac{1}{2}\\-2&1\end{array}\right]$