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garik1379 [7]
3 years ago
6

7. Which dimensions result in the minimum perimeter for a rectangle with area 42.0 cm2?

Mathematics
1 answer:
natali 33 [55]3 years ago
8 0

9514 1404 393

Answer:

  7. d. l = 6.48 cm, w = 6.48 cm

  8. d. square

Step-by-step explanation:

For a given area, the perimeter can always be shortened by reducing the length of the long side and increasing the length of the short side. When you get to the point where you can't do that, then you have the minimum perimeter. You will reach that point when the sides are the same length: the rectangle is a square.

__

7. In light of the above, the best dimensions are √42 ≈ 6.48 cm for length and width.

__

8. In light of the above, the shape is a square.

_____

The attached graph shows the length of one side (x) and the associated perimeter. The other side is 42/x, which will also be 6.48.

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A is true but f(x) is wrong need help
MrRissso [65]

Answer:

f(x)=-18x^2

Step-by-step explanation:

Given:

1+Integral(f(t)/t^6, t=a..x)=6x^-3

Let's get rid of integral by differentiating both sides.

Using fundamental of calculus and power rule(integration):

0+f(x)/x^6=-18x^-4

Additive Identity property applied:

f(x)/x^6=-18x^-4

Multiply both sides by x^6:

f(x)=-18x^-4×x^6

Power rule (exponents) applied"

f(x)=-18x^2

Check:

1+Integral(-18t^2/t^6, t=a..x)=6x^-3

1+Integral(-18t^-4, t=a..x)=6x^-3

1+(-18t^-3/-3, t=a..x)=6x^-3

1+(6t^-3, t=a..x)=6x^-3

That looks great since those powers are the same on both side after integration.

Plug in limits:

1+(6x^-3-6a^-3)=6x^-3

We need 1-6a^-3=0 so that the equation holds true for all x.

Subtract 1 on both sides:

-6a^-3=-1

Divide both sides by-6:

a^-3=1/6

Raise both sides to -1/3 power:

a=(1/6)^(-1/3)

Negative exponent just refers to reciprocal of our base:

a=6^(1/3)

3 0
3 years ago
I have 15 overdue lessons in class. If each lesson takes 30min to and hour to do and I have 7-8 hours how many lessons could I g
Leokris [45]
About 4 if each is 30 mins
lol...
5 0
3 years ago
Evaluate the given integral by changing to polar coordinates. sin(x2 + y2) dA R , where R is the region in the first quadrant be
koban [17]

Set

\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\implies\mathrm dA=r\,\mathrm dr\,\mathrm d\theta

The region R is given in polar coordinates by the set

R=\left\{(r,\theta)\mid2\le r\le3,0\le\theta\le\dfrac\pi2\right\}

So we have

\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_0^{\pi/2}\int_2^3r\sin(r^2)\,\mathrm dr\,\mathrm d\theta=\boxed{\frac\pi4(\cos4-\cos9)}

8 0
3 years ago
Simplify the rational expression:<br> 5y+5/2 ÷ 25y-20/40y²-32y
NNADVOKAT [17]

Step-by-step explanation:

Your problem → 5y+5/2 / 25y-20/40y^2-32y

5y+5/2÷25y-20/40y^2-32y

=5⋅y+5/2÷25×y-20/40×y^2-32×y

=5y+5/2÷25×y-20/40×y^2-32y

=5y+5/2×1/25×y-20/40×y^2-32y

=5y+y/10-20/40×y^2-32y

=5y+y/10-y^2/2-32y

=5y×10+y-y^2×5-32y×10 ÷ 10

=50y+y-5y2-320y ÷ 10

= -269y-5y^2 / 10

4 0
3 years ago
Can someone help me please???​
Oduvanchick [21]

Answer:

1. CHECK

2. CHECK

Step-by-step explanation:

i hole it helps

8 0
3 years ago
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