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GuDViN [60]
4 years ago
7

A rectangle has a perimeter of 32 in. find the length and width of the rectangle under which the area is the largest. follow the

steps: (a) let the width to be x and the length to be y , then the quantity to be maximized is (expressed as a function of both x
Mathematics
1 answer:
Dafna1 [17]4 years ago
5 0
Let width and length be x and y respectively.
Perimeter (32in) =2x+2y=> 16=x+y => y=16-x
Area, A = xy = x(16-x) = 16x-x^2
The function to maximize is area: A=16 x-x^2
For maximum area, the first derivative of A =0 => A'=16-2x =0
Solving for x: 16-2x=0 =>2x=16 => x=8 in
And therefore, y=16-8 = 8 in
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A parallelogram has a base of 16 centimeters and a height of 12 centimeters. What is the area of the parallelogram?
Tatiana [17]

Answer:

Correct (192)

Step-by-step explanation:

The formula for a parallelogram's area is just base times height. 16x12 is equal to 192, so you are correct.

8 0
2 years ago
Read 2 more answers
In the right triangle shown, DF=EF=3DF=EF=3D, F, equals, E, F, equals, 3.
timofeeve [1]

Answer:

DE = 4.24

Step-by-step explanation:

The diagram is not shown; However, the question can still be solved without the diagram.

Having said that,.

Given that

DF = EF = 3

This implies that DE is the hypothenus and will be solved using Pythagoras theorem.

DE² = EF² + DF²

Substitute 3 for EF and DF

DE² = 3² + 3²

DE² = 9 + 9

DE² = 18

Take Square Root of both sides

DE = √18

DE = 4.242640687119285

DE = 4.24 (Approximated)

8 0
4 years ago
It takes 5 hours for a messenger to reach its destination at a speed of 42 mph. If you want to make the journey in 3 and a half
Viktor [21]

Answer:

speed =60 mph

Step-by-step explanation:

speed = distance / time

=( 42 \times 5) \div 3.5 \\  = 60 \\

5 0
4 years ago
Can someone please help?
puteri [66]

Answer:

1)Alternate exterior

2) corresponding

3) alternate interior

4) non of these

Mark me brainliest

6 0
4 years ago
Use Green's Theorem to evaluate F · dr. C (Check the orientation of the curve before applying the theorem.)F(x, y) = y cos(x) −
Sergeeva-Olga [200]

Notice that <em>C</em> has a clockwise orientation. By Green's theorem, we have

\displaystyle\int_C\mathbf F(x,y)\cdot\mathrm d\mathbf r=-\iint_D\left(\frac{\partial(xy+x\cos x)}{\partial x}-\frac{\partial(y\cos x-xy)}{\partial y}\right)\,\mathrm dx\,\mathrm dy

where <em>D</em> is the triangule region with <em>C</em> as its boundary, given by the set

D=\{(x,y)\mid0\le x\le2\land0\le y\le8-4x\}

So we have

\displaystyle\int_C\mathbf F(x,y)\cdot\mathrm d\mathbf r=-\int_0^2\int_0^{8-4x}((y+\cos x-x\sin x)-(\cos x-x\sin x))\,\mathrm dy\,\mathrm dx

\displaystyle\int_C\mathbf F(x,y)\cdot\mathrm d\mathbf r=-\int_0^2\int_0^{8-4x}y\,\mathrm dy\,\mathrm dx=\boxed{-\dfrac{64}3}

5 0
3 years ago
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