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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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Equation equivalent to 5x(3x+4)
Vika [28.1K]

use the distributive property

5x(3x+4)

15x^2 +20x

7 0
3 years ago
Read 2 more answers
2.4x-9 < 1.8x+6 solve the inequality
nasty-shy [4]

Answer:

x<25

Step-by-step explanation:

Let's solve your inequality step-by-step.

2.4x−9<1.8x+6

Step 1: Subtract 1.8x from both sides.

2.4x−9−1.8x<1.8x+6−1.8x

0.6x−9<6

Step 2: Add 9 to both sides.

0.6x−9+9<6+9

0.6x<15

Step 3: Divide both sides by 0.6

0.6x/0.6 < 15/0.6

x<25

8 0
3 years ago
Please!!! I NEED help with this!!!! it's due in an hour and I have no idea how to do it!!!
sergejj [24]

Answer:

Well you can try to use a calculator, like the one on google. I can try to help if you really need it.

Step-by-step explanation:


3 0
3 years ago
Researchers recorded the speed of ants on trails in their natural environments. The ants studied, Leptogenys processionalis, all
posledela

This question is Incomplete

Complete Question

Researchers recorded the speed of ants on trails in their natural environments. The ants studied, Leptogenys processionalis, all have the same body size in their adult phase, which made it easy to measure speeds in units of body lengths per second (bl/s). The researchers found that, when traffic is light and not congested, ant speeds vary roughly Normally, with mean 6.20 bl/s and standard deviation 1.58 bl/s. (a) What is the probability that an ant's speed in light traffic is faster than 5 bl/s? You may find Table B useful. (Enter your answer rounded to four decimal places.)

Answer:

0.7762

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

Population mean = 6.20 bl/s

Standard deviation = 1.58 bl/s.

x = 5 bl/s

z = 5 - 6.20/1.58

z = -0.75949

The probability that an ant's speed in light traffic is faster than 5 bl/s is P( x > 5)

Probability value from Z-Table:

P(x<5) = 0.22378

P(x>5) = 1 - P(x<5)

= 1 - 22378

= 0.77622

Approximately to 4 decimal places = 0.7762

The probability that an ant's speed in light traffic is faster than 5 bl/s is 0.7762

5 0
3 years ago
Expand 4x(7x-11)<br> For maths <br> Help please
harina [27]

Answer:

28x^2-44x

Step-by-step explanation:

Use the distributive property.

4x * 7x = 28x^2

4x * (-11) = -44x

We will get 28x^2-44x

Hope this helps!

7 0
3 years ago
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