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3 years ago

5

In a greenhouse, basil is grown in planters that have a base of 4 1 4 in by 1 2 3 in. If there is a 4 × 4 array of these planter

s with no space between them, what is the area covered by the planters?

1 answer:

Kryger [21]3 years ago

8 0

Answer:

The area covered by the planters is 113.6 square inches.

Step-by-step explanation:

Assuming that the dimmensions of the greenhouse are 4 1/4 by 1 2/3 inches. We first need to convert them from mixed fractions to normal fractions, we do that by summing the integer part to the fraction one. We have:

length = 4 + 1/4 = 16/4 + 1/4 = 17/4 inches

width = 1 + 2/3 = 3/3 + 2/3 = 5/3 inches

The area of the greenhouse is given by:

area = length*width

area = (17/4)*(5/3) = 85/12 = 7.1 square inches

Since there is an array of 4x4 there is a total of 16 greenhouses and it's total area is 16*7.1 = 113.6 square inches

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Tom opened a lemonade stand to earn money to buy a new bike for $200. He sells a cup of lemonade for $1.00. On the first day he

tia_tia [17]

1 - 8

2 - 16

3 - 32

x - +200

Each lemonade has a profit of $1, so the number of cups of lemonade he sells is the same as the amount of dollars he has. You just need to add each previous day's total to the next day's, doubling the amount of cups from the previous day's for each new day.

4 - (32 x 2 = ) 64

8 + 16 + 32 + 64 = 120. We need at least $80 dollars more, so we continue.

5 - (64 x 2 = ) 128

120 + 128 = $248.00 That's more than enough, so Tom will have enough money for his bike after the fifth day of selling lemonade.

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3 years ago

A water balloon is catapulted into the air so that it's height, in meters, after t seconds is h = -4.9t^ +27t +2.4

zysi [14]

$h = -4..9t^2+27t+2.4\ for\ t=3$

Just replace t with 3 and solve.

$h = -4.9*3^2 + 27*3 + 2.4 \\ h = -4.9*9 + 27*3 + 2.4 \\ h=-44.1 + 81 + 2.4 \\ \boxed{h=39.3}$

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3 years ago

Find all the zeroes of the equation(with simple steps).

uysha [10]

<u>Answer-</u>

<em>The zeros are, $5,\ -5,\ 4i,\ -4i$ </em>

<u>Solution-</u>

$\Rightarrow -3x^4+27x^2+1200=0$

$\Rightarrow -3(x^2)^2+27(x^2)+1200=0$

Here,

a = -3, b = 27, c = 1200

So,

$x^2=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\dfrac{-27\pm \sqrt{-27^2-4\cdot (-3)\cdot 1200}}{2\cdot (-3)}$

$=\dfrac{-27\pm \sqrt{729+14400}}{-6}$

$=\dfrac{-27\pm 123}{-6}$

$=\dfrac{-27+123}{-6},\ \dfrac{-27- 123}{-6}$

$=\dfrac{96}{-6},\ \dfrac{-150}{-6}$

$=-16,\ 25$

So,

$\Rightarrow x^2=25,\ -16$

$\Rightarrow x=\sqrt{25},\ \sqrt{-16}$

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3 years ago

Find the exact value of cot(15degrees) using half angle formulas?

liubo4ka [24]

Double angle (or half angle, depending how you look at it) identities:

$\cos^2\dfrac x2=\dfrac{1+\cos x}2$

$\sin^2\dfrac x2=\dfrac{1-\cos x}2$

$\implies\cot^2\dfrac x2=\dfrac{\cos^2\frac x2}{\sin^2\frac x2}=\dfrac{1+\cos x}{1-\cos x}$

So we have

$\cot^215^\circ=\dfrac{1+\cos30^\circ}{1-\cos30^\circ}=\dfrac{1+\frac{\sqrt3}2}{1-\frac{\sqrt3}2}$

$\implies\cot^215^\circ=7+4\sqrt3$

$\implies\cot15^\circ=\sqrt{7+4\sqrt3}=2+\sqrt3$

Note that when taking the square root, we should take into account that that could yield two possible solutions, but we know $\cos15^\circ>0$ and $\sin15^\circ>0$ , so it's also the case that $\cot15^\circ>0$ .

Also, the reason we have equality in the last step can be explained like so:

$7+4\sqrt3=4+4\sqrt3+3=4+4\sqrt3+(\sqrt3)^2=(2+\sqrt3)^2$

(not unlike the process used to complete the square)

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3 years ago

Read 2 more answers

A survey team is trying to estimate the height of a mountain above a level plain. From one point on the plain, they observe that

PilotLPTM [1.2K]

The tangent or tanθ in a right-angle triangle is the ratio of its perpendicular to its base. The height of the mountain is 3110.5767 feet.

<h3>What is Tangent (Tanθ)?</h3>

The tangent or tanθ in a right-angle triangle is the ratio of its perpendicular to its base. it is given as,

$\rm Tangent(\theta) = \dfrac{Perpendicular}{Base}$

where,

θ is the angle,

Perpendicular is the side of the triangle opposite to the angle θ,

The base is the adjacent smaller side of the angle θ.

The height of the mountain from the first point,

$\tan(29^o) = \dfrac{H}{x+1000}\\\\H = (1000+x) \times \tan(29^o)$

The height of the mountain from the second point,

$\tan(34^o) = \dfrac{H}{x}\\\\H = x \times \tan(34^o)$

Equating the height,

H = H

(1000+x) × tan(29°) = x tan( 34°)

x = 4611.5767 feet

Now, the height of the mountain is,

H = x tan(34°)

H = 3110.54776 feet

Hence, the height of the mountain is 3110.5767 feet.

Learn more about Tangent (Tanθ):

brainly.com/question/10623976

#SPJ1

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2 years ago