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irinina [24]
3 years ago
7

If the area around the vegetable garden is of uniform width (labeled with x) and the dimensions of the vegetable garden is 45 fe

et by 20 feet, what expression represents the area of the flower garden?
Make sure to show all of your steps in your answer, including the area of the vegetable garden and the area of the entire garden.

Mathematics
1 answer:
Nady [450]3 years ago
7 0
Area of Rectangle = Width × Length

Vegetable Garden
Width = 20
Length = 45
Area of Vegetable Garden = 20×45=900 Sq ft

Area of Entire Garden = Width × Length
Width = 20 + 2x
Length = 45 + 2x
Area = (20 + 2x)(45 + 2x)
(2x + 20)(2x + 45)
4x^2 + 90x + 45x + 900
Area of Entire Garden = 4x^2 + 135x + 900

Area of Flower Garden = Area of Entire Garden - Area of Vegetable Garden

(4x^2 + 135x + 900) - (20 × 45)
4x^2 + 135x + 900 - 900
Area of Flower Garden = 4x^2 + 135x

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Noel and Casey both start at the same place. Noel walks due south and Casey walks due east. After some time has passed, Noel is
Dovator [93]

Answer:

2.04 miles per hour

Step-by-step explanation:

Given

Noel

n_1 =6miles

r_1 = 2mph

Casey

c_1 = 8miles

r_2 =1mph

Required

The rate at which the distance increases

Their movement forms a right triangle and the distance between them is the hypotenuse.

At n_1 =6miles and c_1 = 8miles

The distance between them is:

d_1 = \sqrt{n_1^2 + c_1^2}

d_1 = \sqrt{6^2 + 8^2}

d_1 = \sqrt{100}

d_1 = 10miles

After 1 hour, their new position is:

New = Old + Rate * Time

n_2 = n_1 + r_1 * 1

n_2 = 6 + 2 * 1 = 8

And:

c_2 = c_1 + r_2 * 1

c_2 = 8 + 1 * 1 = 9

So, the distance between them is now:

d_2 = \sqrt{n_2^2 + c_2^2}

d_2 = \sqrt{8^2 + 9^2}

d_2 = \sqrt{145}

d_2 = 12.04

The rate of change is:

\triangle d = d_2 -d_1

\triangle d = 12.04 -10

\triangle d = 2.04

5 0
3 years ago
Find all numbers whose absolute value is -9.
Neko [114]

Answer:

None

Step-by-step explanation:

A number's absolute value will always be positive

|9| = 9

|-9| = 9

3 0
3 years ago
Read 2 more answers
Which best explains how you know that P(2, 7) is on the same line as Q(7, 4) and R(12, 1)?
Helen [10]

Answer:

See explanation below

Step-by-step explanation:

The best explanation is noticing that in order to get from the point R (12, 1) to the point Q (7, 4) we move 5 units to the left and 3 units up. And to go from point Q (7, 4) to point P (2, 7) we do exactly the same: move 5 units to the left and 3 units up. That means that these points are all connected via the same rate of change:  - 3/5, which is in fact the slope of the line the three points belong to.

4 0
3 years ago
Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat
zavuch27 [327]

Answer:

The equation contains exact roots at x = -4 and x = -1.

See attached image for the graph.

Step-by-step explanation:

We start by noticing that the expression on the left of the equal sign is a quadratic with leading term x^2, which means that its graph shows branches going up. Therefore:

1) if its vertex is ON the x axis, there would be one solution (root) to the equation.

2) if its vertex is below the x-axis, it is forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will not have real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently:

We recall that the x-position of the vertex for a quadratic function of the form f(x)=ax^2+bx+c is given by the expression: x_v=\frac{-b}{2a}

Since in our case a=1 and b=5, we get that the x-position of the vertex is: x_v=\frac{-b}{2a} \\x_v=\frac{-5}{2(1)}\\x_v=-\frac{5}{2}

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:

y_v=f(-\frac{5}{2})\\y_v=(-\frac{5}{2} )^2+5(-\frac{5}{2} )+4\\y_v=\frac{25}{4} -\frac{25}{2} +4\\\\y_v=\frac{25}{4} -\frac{50}{4}+\frac{16}{4} \\y_v=-\frac{9}{4}

This is a negative value, which points us to the case in which there must be two real solutions to the equation (two x-axis crossings of the parabola's branches).

We can now continue plotting different parabola's points, by selecting x-values to the right and to the left of the x_v=-\frac{5}{2}. Like for example x = -2 and x = -1 (moving towards the right) , and x = -3 and x = -4 (moving towards the left.

When evaluating the function at these points, we notice that two of them render zero (which indicates they are the actual roots of the equation):

f(-1) = (-1)^2+5(-1)+4= 1-5+4 = 0\\f(-4)=(-4)^2+5(-4)_4=16-20+4=0

The actual graph we can complete with this info is shown in the image attached, where the actual roots (x-axis crossings) are pictured in red.

Then, the two roots are: x = -1 and x = -4.

5 0
3 years ago
A T-shirt is on sale for 20 % off. The sale price is $18 ,What was the original price?Explain.
Pavel [41]
The original price was $22.5.
18 divided by 0.8 is 22.5.
If you double check 22.5 times 0.8, you will get 18, your sales price.
6 0
3 years ago
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