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tensa zangetsu [6.8K]

3 years ago

12

For a party, Jon is making sandwiches that are 1 1/4 feet long. He buys 7/8 pound of ham and 1 3/8 pounds of turkey for the sand

wiches. Then he puts 1/6 of the meat on each sandwich. How many pounds of meat does Jon put on each sandwich?

1 answer:

Vinil7 [7]3 years ago

8 0

Answer:

3/8 pounds of meat

Step-by-step explanation:

7/8 pounds of ham

= 0.875 pounds of ham

1 3/8 pounds of turkey

= 11/8 pounds of turkey

= 1.375 pounds of turkey

Pounds of meat

P_meat = 0.875 pounds of ham + 1.375 pounds of turkey

P_meat = 2.25 pounds of meat

Jon puts 1/6 of the meat on each sandwich

P_meat/6 = 3/8 pounds = 0.375 pounds

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B, the bottom of the figure contains points E, F, and H

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Step-by-step explanation:

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Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l

SIZIF [17.4K]

Answer:

$\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}$

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let

$C_1,C_2$

be the two paths.

Recall that if we parametrize a path C as $(r_1(t),r_2(t),r_3(t))$ with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

$\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt$

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and

$\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}$

The parametrization of the line segment from (2,2,2) to

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)

r'(t) = (-11,4,3)

and

$\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}$

Hence

$\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}$

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10| x – 4| - 3 > 47

Andrew [12]

Answer:

x < -1 or x > 9

Step-by-step explanation:

$10|x-4|-3>47$

Add 3 to both sides.

$10|x-4|-3+3>47+3$

Simplify.

$10|x-4|>50$

Divide both sides by 10.

$\frac{10|x-4|}{10} >\frac{50}{10}$

Simplify.

$|x-4|>5$

Apply absolute value rule: If $|u|>a,a>0$ then $u < -a$ or $u>a$

$x-4$ or $x-4 >5$

$x-4$

Add 4 to both sides.

$x-4+4$

Simplify.

$x$

$x-4>5$

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$x-4+4>5+4$

Simplify.

$x>9$

Combine the intervals.

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3 years ago