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kotykmax [81]
3 years ago
9

Please help me solve this 1/8 in.= 10 mi. 1 1/8 in.= ____ mi.

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
3 0

Answer:

2.1701x10^-5 mi

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i accidentaly put awnser and i dont know how to get out of here

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I would love your guys help!! ASAP!!
sweet [91]

9514 1404 393

Answer:

  1. 3/2
  2. it is the ratio of oil to vinegar in the dressing
  3. y = 3/2x

Step-by-step explanation:

1. The table can be filled in like this:

  • (x, y)
  • (0, 0)
  • (2, 3)
  • (4, 6)
  • (6, 9)

The constant of proportionality is the ratio of y to x:

  k = 3/2

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2. The meaning of the constant of proportionality is just what it is defined to mean: the ratio of oil to vinegar in the dressing.

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3. The equation is y=kx, where k is the value in part 1:

  y = (3/2)x

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3 years ago
Which expression is equivalent to 52(2x+1)?
Mama L [17]

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52 x 2 + 1 - 1 = 104

Adding some description to extend my answer since the equation was too short.

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3 years ago
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If X = {−1, 0, 2} and Y = {1, 2, 5}, find X ∪ Y.
malfutka [58]

Answer:

(0,-1,2,5)

Step-by-step explanation:

X = (-1, 0, 2)\\Y = (1, 2, 5)\\ \\X U Y = ?\\= (-1, 0,2,5)

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3 years ago
99 POINT QUESTION, PLUS BRAINLIEST!!!
Elden [556K]
We draw region ABC. Lines that connect y = 0 and y = x³ are vertical so:
(i) prependicular to the axis x - disc method;
(ii) parallel to the axis y - shell method;
(iii) parallel to the line x = 18 - shell method.

Limits of integration for x are easy x₁ = 0 and x₂ = 9.
Now, we have all information, so we could calculate volume.

(i)

V=\pi\cdot\int\limits_a^bf^2(x)\, dx\qquad\implies \qquad a=0\qquad b=9\qquad f(x)=x^3


V=\pi\cdot\int\limits_0^9(x^3)^2\, dx=\pi\cdot\int\limits_0^9x^6\, dx=\pi\cdot\left[\dfrac{x^7}{7}\right]_0^9=\pi\cdot\left(\dfrac{9^7}{7}-\dfrac{0^7}{7}\right)=\dfrac{9^7}{7}\pi=\\\\\\=\boxed{\dfrac{4782969}{7}\pi}

Answer B. or D.

(ii)

V=2\pi\cdot\int\limits_a^bx\cdot f(x)\, dx


V=2\pi\cdot\int\limits_0^{9}(x\cdot x^3)\, dx=2\pi\cdot\int\limits_0^{9}x^4\, dx=
2\pi\cdot\left[\dfrac{x^5}{5}\right]_0^9=2\pi\cdot\left(\dfrac{9^5}{5}-\dfrac{0^5}{5}\right)=\\\\\\=2\pi\cdot\dfrac{9^5}{5}=\boxed{\dfrac{118098}{5}\pi}

So we know that the correct answer is D.

(iii)
Line x = h

V=2\pi\cdot\int\limits_a^b(h-x)\cdot f(x)\, dx\qquad\implies\qquad h=18


V=2\pi\cdot\int\limits_0^9\big((18-x)\cdot x^3\big)\, dx=2\pi\cdot\int\limits_0^9(18x^3-x^4)\, dx=\\\\\\=2\pi\cdot\left(\int\limits_0^918x^3\, dx-\int\limits_0^9x^4\, dx\right)=2\pi\cdot\left(18\int\limits_0^9x^3\, dx-\int\limits_0^9x^4\, dx\right)=\\\\\\=2\pi\cdot\left(18\left[\dfrac{x^4}{4}\right]_0^9-\left[\dfrac{x^5}{5}\right]_0^9\right)=2\pi\cdot\Biggl(18\biggl(\dfrac{9^4}{4}-\dfrac{0^4}{4}\biggr)-\biggl(\dfrac{9^5}{5}-\dfrac{0^5}{5}\biggr)\Biggr)=\\\\\\

=2\pi\cdot\left(18\cdot\dfrac{9^4}{4}-\dfrac{9^5}{5}\right)=2\pi\cdot\dfrac{177147}{10}=\boxed{\dfrac{177147\pi}{5}}

Answer D. just as before.

6 0
3 years ago
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