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Norma-Jean [14]
2 years ago
13

Can someone help me plseawe

Mathematics
2 answers:
Tems11 [23]2 years ago
7 0
Here’s is your answer

Georgia [21]2 years ago
3 0

18 meters by 14 meters

Step-by-step explanation:

18 meters by 14 meters

Area of rectangle is base times height (AKA length times width or however you wish to call the dimensions).

This question is a system of equations question. We can make two equations. We know that every rectangle must have 2 "lengths" and 2 "widths" (4 sides in a rectangle). Since Maria only has 64 meters of fencing, we get the equation:

2L+2W = 64

Now we know the area enclosed must be 252 so we have:

L* W = 252

Rewriting the second equation we get:

Plug this into first equation to get:

(multiply everything by L)

Solving the quadratic tells us L can be 18 or 14. Let's consider if L is 18 first by plugging 18 back into the second equation to get 18 * W = 252 which gives us W = 14. If we do the same for when L = 14, we get that W = 18.

Now you might wonder why we have 2 possible combinations, that's because in reality, it doesn't matter which one we label as length (L) or width (W).

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Answer: P(x) = {(x-4)^2} (x) (x+4)

Step-by-step explanation:

Let's start with the multiplicity of 2;

At multiplicity of 2; x=4.

Therefore, x - 4 is a factor of the function P(x).

Since it has a multiplicity of 2, we will rewrite the factor as (x-4)^2

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At this multiplicity of 1, x= 0 and - 4.

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Since multiplicity of 1, the factors remain as they are without any additional root on top.

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The graph below shows the average Valentine’s Day spending between 2003 and 2012
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The average rate of change of a graph between two intervals is given by the difference in value of the values on the graph of the two interval divided by the difference between the two intervals.

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From the graph the average Valentine's day spending in 2005 is 98 while the average Valentine's day spending in 2007 is 120.

The average rate of change in spending between 2005 and 2007 is given by

\frac{120-98}{2007-2005} = \frac{22}{2} =\$11/year



Part B

From the graph the average Valentine's day spending in 2004 is 100 while the average Valentine's day spending in 2010 is 103.

The average rate of change in spending between 2004 and 2010 is given by

\frac{103-100}{2010-2004} = \frac{3}{6} =\$0.5/year



Part C:

From the graph the average Valentine's day spending in 2009 is 102 while the average Valentine's day spending in 2010 is 103.

The average rate of change in spending between 2009 and 2010 is given by

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4 years ago
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