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3 years ago

The area of a rectangle is 53.3 in2. If the width is 6.5 in, what is the perimeter of the rectangle?

2 answers:

7 0

You can find the length from
A = LW
53.3 in² = L×(6.5 in)
(53.3 in²)/(6.5 in) = L = 8.2 in

The perimeter is twice the sum of length and width.
P = 2(L +W)
P = 2(8.2 in +6.5 in) = 29.4 in

The perimeter of the rectangle is 29.4 inches.

elixir [45]3 years ago

6 0

Area of a rectangle formula
l × w
6.5 is our width you already know but 53.3 isn't the length so really
you are trying to solve for l, so l × 6.5 = 53.3
when you get that answer you should get 8.2

Now .. you already have the width and the length so you hook in 6.5(w) + 6.5(w)+ 8.2 (l)+ 8.2(l)
6.5 + 6.5 gives you 13 and
8.2 + 8.2 gives you 16.4
Adding both up you get the final answer/ perimeter 29.4 inches

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3 years ago

Subtract and add the group of fractions below. Make sure the answer is fully reduced.

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3 years ago

Read 2 more answers

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ELEN [110]

For the first one c-9=(2)(4)+6
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Third, add 4 to both sides. a=10.
Fourth, subtract 8 from both sides. d=-9
Fifth, 5+2(3+4)=-x.
First we are going to add 3 and 4. 3+4=7. 5+2(7)=-x. Next we are going to multiply 2 and 7. 2*7=14. 5+14=-x. 5+14=19. So our equation is 19=-x. Next we are going to multiply both sides by -1. Making our final equation x=-19.
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3 years ago

Consider the line integral Z C (sin x dx + cos y dy), where C consists of the top part of the circle x 2 + y 2 = 1 from (1, 0) t

pashok25 [27]

Direct computation:

Parameterize the top part of the circle $x^2+y^2=1$ by

$\vec r(t)=(x(t),y(t))=(\cos t,\sin t)$

with $0\le t\le\pi$ , and the line segment by

$\vec s(t)=(1-t)(-1,0)+t(2,-\pi)=(3t-1,-\pi t)$

with $0\le t\le1$ . Then

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)$

$=\displaystyle\int_0^\pi(-\sin t\sin(\cos t)+\cos t\cos(\sin t)\,\mathrm dt+\int_0^1(3\sin(3t-1)-\pi\cos(-\pi t))\,\mathrm dt$

$=0+(\cos1-\cos2)=\boxed{\cos1-\cos2}$

Using the fundamental theorem of calculus:

The integral can be written as

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=\int_C\underbrace{(\sin x,\cos y)}_{\vec F}\cdot\underbrace{(\mathrm dx,\mathrm dy)}_{\vec r}$

If there happens to be a scalar function $f$ such that $\vec F=\nabla f$ , then $\vec F$ is conservative and the integral is path-independent, so we only need to worry about the value of $f$ at the path's endpoints.

This requires

$\dfrac{\partial f}{\partial x}=\sin x\implies f(x,y)=-\cos x+g(y)$

$\dfrac{\partial f}{\partial y}=\cos y=\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=\sin y+C$

So we have

$f(x,y)=-\cos x+\sin y+C$

which means $\vec F$ is indeed conservative. By the fundamental theorem, we have

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=f(2,-\pi)-f(1,0)=-\cos2-(-\cos1)=\boxed{\cos1-\cos2}$

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3 years ago

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dybincka [34]

2x/7 = 12/14
Multiply both sides of the equation by 7:
2x = 84/14
Simplify:
2x = 6
Divide both sides by 2 and your answer is....
x = 3

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3 years ago