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3 years ago

-16 = -5a -21. Plz help. Show work if possible. Pre-Algebra is hard. :(

Mathematics

1 answer:

Leona [35]3 years ago

5 0

Answer:

a = -1

Step-by-step explanation:

-16 = -5a - 21 (Given)

5 = -5a (Added 21 on both sides)

-1 = a (Divided -5 on both sides)

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The length of a rectangle is 5 centimeters less then twice it's width. The perimeter of the rectangle is 26 cm. What are the dim

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Answer:

The dimensions of the rectangle are length = 7cm and width = 6cm.

Step-by-step explanation:

In order to solve for the dimensions, you will need to set up two equations in order to solve for the missing variable. Given the information that the length is 5 cm less then twice it's width, using 'L' for length and 'w' for width we get the following equation: L = 2w - 5. Perimeter is the sum of all the sides, or in the case of a rectangle P = 2w + 2L. We can then use our expression for 'L' in our perimeter formula: 26 = 2w + 2(2w - 5). First, using the distributive property we get: 26 = 2w + 4w - 10. Next, we combine like terms: 26 = 6w - 10. Then, we use inverse operations to isolate the variable: 26 + 10 = 6w - 10 + 10 to get 36 = 6w, divide both sides by 6 to get w = 6. Lastly, plug in the value of 'w' to 'L': L = 2(6) - 5 or L = 7.

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3 years ago

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ANSWER:

The average burnout time of a large number of bulbs has a sampling distribution that is close to Normal.

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3 years ago

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2 years ago

Evaluate the surface integral. s x2 + y2 + z2 ds s is the part of the cylinder x2 + y2 = 4 that lies between the planes z = 0 an

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Parameterize the lateral face $T_1$ of the cylinder by

$\mathbf r_1(u,v)=(x(u,v),y(u,v),z(u,v))=(2\cos u,2\sin u,v$

where $0\le u\le2\pi$ and $0\le v\le3$ , and parameterize the disks $T_2,T_3$ as

$\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)$
$\mathbf r_3(r,\theta)=(r\cos\theta,r\sin\theta,3)$

where $0\le r\le2$ and $0\le\theta\le2\pi$ .

The integral along the surface of the cylinder (with outward/positive orientation) is then

$\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}((2\cos u)^2+(2\sin u)^2+v^2)\left\|{{\mathbf r}_1}_u\times{{\mathbf r}_2}_v\right\|\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+0^2)\left\|{{\mathbf r}_2}_r\times{{\mathbf r}_2}_\theta\right\|\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+3^2)\left\|{{\mathbf r}_3}_r\times{{\mathbf r}_3}_\theta\right\|\,\mathrm d\theta\,\mathrm dr$
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$=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr$
$=136\pi$

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3 years ago

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V125BC [204]

If $a=48Tg\pi$ then $g=\frac{a}{48T\pi}$ .

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3 years ago