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4 years ago

6

Oliver is building a box kite. The frame of the kite has the dimensions shown. He covered the sides of the kite (but not the top

and bottom) with 720072007200 cm^2 2 start superscript, 2, end superscript of lightweight fabric.

1 answer:

Mamont248 [21]4 years ago

5 0

Answer:

h=60 and v=54000 cm³.

Step-by-step explanation:

Let h be the height of the box kite,

Assume that the length of one face of base is 30 cm and another face of the base is 30 cm.

The total area of the kite box that is covered by the fabric

$Area= 4 \times \text{Area of one face}$

$Area= 4 \times (30\times h)$

$Area= 120h$

The area of 4 sides of the kite is 7200.

$120h = 7200$

Divide both sides by 120.

$h=60$

The height of the box is 60 cm.

The volume of a rectangular prism.

$V=length\times breadth\times height$

$V=30\times 30\times 60$

$V=54000$

Therefore, the height of the kite is 60 units and volume of the volume inside the kite is 54000 cm³.

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3 years ago

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Determine whether this pair of lines is parellel, perpendicular, or neither 6+2x=3y 3x+2y=9 Choose the correct answer below. A.

andriy [413]

Answer:

Option A. These two lines are perpendicular

Step-by-step explanation:

we know that

If two lines are parallel, then their slopes are equal

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

step 1

Convert the given equation A in slope intercept form

$y=mx+b$

where

m is the slope

b is the y-intercept

we have

$6+2x=3y$ ----> equation A

Solve for y

That means----> isolate the variable y

Divide by 3 both sides

$y=\frac{2}{3}x+2$

so

$m_A=\frac{2}{3}$

step 2

Convert the given equation B in slope intercept form

$y=mx+b$

where

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we have

$3x+2y=9$ ----> equation B

Solve for y

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3 0

4 years ago

posledela

Answer:

<h2>(90Degree)(0.3Degree/1min) = 300min</h2><h2>since 60min = 1hr</h2><h2>(300min)(1hr/60min) = 5hrs</h2>

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3 years ago

What is the equation of the line that is parallel to y=1/2x+2

abruzzese [7]

When y =0 or when z =0

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3 years ago

Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = 2 cos(t) + sin(2t),[0, π/2].

castortr0y [4]

Answer:

The absolute maximum is $\frac{3\sqrt 3}2$ and the absolute minimum value is $0.$

Step-by-step explanation:

Differentiate of $f$ both sides w.r.t. $t$ ,

$f(t)=2 \cos t+\sin 2t$

$\Rightarrow f'(t)=-2\sin t+2\cos 2t$

Now take $f'(t)=0$

$\Rightarrow -2\sin t+2\cos 2t=0$

$\Rightarrow 2\cos 2t=2\sin t$

$\Rightarrow \cos 2t=\sin t$

$\Rightarrow 1-2\sin ^2t =\sin t \quad \quad [\because \cos 2t = 1-2\sin ^2t]$

$\Rightarrow 2\sin ^2t+\sin t-1=0$

$\Rightarrow 2\sin ^2t+2\sin t-\sin t-1=0$

$\Rightarrow 2\sin t(\sin t+1)-1(\sin t+1)=0$

$\Rightarrow (\sin t+1)(2\sin t-1)=0$

$\Rightarrow \sin t+1=0 \;\text{and}\; 2\sin t-1=0$

$\Rightarrow \sin t =-1 \;\text{and}\; \sin t =\frac 12$

In the interval $0\leq t\leq \frac {\pi}2$ , the answer to this problem is $\frac {\pi}6$

Now find the second derivative of $f(t)$ w.r.t. $t$ ,

$f''(t)=-2\cos t-4\sin 2t$

$\Rightarrow \left[f''(t)\right]_{t=\frac {\pi}6}=-2\times \frac {\sqrt 3}2-4\times \frac{\sqrt 3}2=-3\sqrt 3$

Thus, $f(t)$ is maximum at $t=\frac {\pi}6$ and minimum at $t=0$

$\left[f(t)\right]_{t=\frac {\pi}6}=2\times \frac {\sqrt 3}2+\frac{\sqrt 3}2=\frac{3\sqrt 3}2\;\text{and}\;\left[f(t)\right]_{t=\frac{\pi}2}= 2\times 0+0=0$

Hence, the absolute maximum is $\frac{3\sqrt 3}2$ and the absolute minimum value is $0$ .

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3 years ago