What is 81/90 in simplest form?

Jake has a rectangular garden that measures 12 feet by 14 feet. He wants to increase the area by 50% and plans to increase each

emmasim [6.3K]

<em>In order to increase the area of a rectangular garden that measures 12 feet by 14 feet by 50% Jake must increase each dimension by equal lengths, x:</em>

$x\approx 2.9ft$

<h2>Explanation:</h2><h2 />

First of all, let's calculate the area of the original rectangular garden:

$A=b\times h \\ \\ b:base \\ \\ h:height \\ \\ \\ b=12ft \\ \\ h=14ft \\ \\ \\ A=12(14) \\ \\ A=168ft^2$

Jake wants to increase the area by 50%, so the new area would be:

$A'=168(1.5) \\ \\ A'=252ft^2$

He wants to increase the area by 50% and plans to increase each dimension by equal lengths, x, so this is represented by the figure below, therefore:

$(12+x)(14+x)=252 \\ \\ 168+12x+14x+x^2-252=0 \\ \\ x^2+26x-84=0 \\ \\ \\ Using \ quadratic \ formula: \\ \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ a=1 \\ \\ b=26 \\ \\ c=-84 \\ \\ \\ x=\frac{-26 \pm \sqrt{26^2-4(1)(-84)}}{2(1)} \\ \\ x=\frac{-26 \pm \sqrt{1012}}{2} \\ \\ \\ Two \ solutions: \\ \\ x_{1}=-13+\sqrt{253} \approx 2.9\\ \\ x_{2}=-13-\sqrt{253} \approx -28.9 \\ \\ x_{2} \ is \ discarded \ because \ it \ can't \ be \ negatives$

Finally:

<em>In order to increase the area of a rectangular garden that measures 12 feet by 14 feet by 50% Jake must increase each dimension by equal lengths, x:</em>

$x\approx 2.9ft$

<h2>Learn more:</h2>

Dilation: brainly.com/question/10945890

#LearnWithBrainly

6 0

3 years ago

Please help me with both questions!!!!

Naya [18.7K]

Answer:

Step-by-step explanation:

16.

points are (0,-3),(2,1),(4,-3)

eq. of line through (0,-3) and (2,1) is

y+3=\frac{1+3}{2-0}(x-0)

or y+3=2x

2x-y=3

as the line is dotted so either < or >.

consider 2x-y>3

put x=0,y=0

0>3

which is not possible .

(0,0) does not satisfy the inequality

Hence shaded region is the required region.

now eq of line through (2,1)and (4,-3) is

y-1=\frac{-3-1}{4-2}(x-2)

y-1=-2(x-2)

y-1=-2x+4

2x+y=5

it is also a dotted line so either < or >

consider 2x+y<5

put x=0,y=0

0<5

which is true.

so (0,0) satisfy this inequality.

so the two inequalities are

2x-y>3

and 2x+y<5

17.

consider the points (0,4),(2,3),(4,4)

eq. of line through (0,4) and (2,3) is

y-4=\frac{3-4}{2-0}(x-0)

y-4=-1/2(x)

2y-8=-x

or x+2y=8

as the line is solid

so either≤ or ≥

consider x+2y≥8

put x=0,y=0

0≥8

which is impossible.

(0,0) does not satisfy the graph.

which is true as graph lies above the line.

again eq. of line through (2,3) and (4,4) is

y-3=\frac{4-3}{4-2}(x-2)

y-3=1/2(x-2)

2y-6=x-2

x-2y=-4

consider x-2y≤-4

put x=0, y=0

0≤-4

which is impossible.

so (0,0) does not satisfy the graph.

so inequality is true as shaded portion is above and left of the line.

so two inequalities are

x+2y≥ 8

and x-2y≤-4

5 0

3 years ago

Which expression is equal to 6(-3m)4?

hichkok12 [17]

Answer: -72m

Step-by-step explanation:

Simplify

7 0

3 years ago

The nth term of a sequence is 2n-1.

Zinaida [17]

Answer:

25,31,37

Step-by-step explanation:

n should be positive integer number. The three numbers in both sequences have different term number n but same value. We can equalize each nth term in the question to "a" which represents one of the three numbers.

a=2n-1, then n=(a+1)/2

a=3n+1, then n=(a-1)/3

remember the two n above are different but both should be positive integer. That means, we have to find the "a" number that gives me an integer n for the first equation. The possible numbers between 20 to 40 are 22,25,28,31,34,37,40.

The possible numbers for the second equation are 21,23,25,27,29,31,33,35,37,39.

Now find the common numbers between the two sets above. They are 25,31,37

6 0

3 years ago

Can I have some help

jok3333 [9.3K]

18/20=27/x
18x=27*20
x=(27*20)/18=30

Answer: 30

6 0

3 years ago