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Setler79 [48]
3 years ago
13

A spherical fish bowl is half-filled with water. T

Mathematics
2 answers:
olganol [36]3 years ago
7 0

The volume of a sphere of radius r is given by

V=\dfrac{4}{3}\pi\cdot r^3

For your given conditions, the volume of 1/2 a sphere becomes

V=\dfrac{1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{22}{7}\cdot 8^3

Seems to match your 2nd choice:

... 1 over 24 over 322 over 7(83)

alisha [4.7K]3 years ago
6 0

its B) 1 over 24 over 322 over 7(82)(16)

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Rank in decreasing order 8.58, 8.508, 7.058, 7.5
8_murik_8 [283]
7.5 8.58 7.058 8.508
8 0
2 years ago
HELP RIGHT NOW AND ILL GIVE BRAINLY IF CORRECT
Stella [2.4K]

Answer:

-9

Step-by-step explanation:

3^{-36}= 27^{-12}

7 0
3 years ago
The total length of pencils A, B and C is 29 cm. Pencil a is 11 cm shorter then pencil B, and pencil B is twice as long a pencil
satela [25.4K]

The length of pencil A is 5 cm

<em><u>Solution:</u></em>

Let the length of pencil A be "x"

Let the length of pencil B be "y"

Let the length of pencil C be "z"

<em><u>The total length of pencils A, B and C is 29 cm</u></em>

Therefore,

length of pencil A + length of pencil B + length of pencil C = 29

x + y + z = 29 ------------ eqn 1

<em><u>Pencil A is 11 cm shorter then pencil B</u></em>

x = y - 11 ------- eqn 2

<em><u>Pencil B is twice as long a pencil C</u></em>

y = 2z

z = \frac{y}{2} ------ eqn 3

<em><u>Substitute eqn 2 and eqn 3 in eqn 1</u></em>

y - 11 + y + \frac{y}{2} = 29\\\\2y + \frac{y}{2} = 29 + 11\\\\\frac{4y+y}{2} = 40\\\\5y = 80\\\\y = 16

<em><u>Substitute y = 16 in eqn 2</u></em>

x = 16 - 11

x = 5

Thus length of pencil A is 5 cm

5 0
3 years ago
Find the exact location of all the relative and absolute extrema of the function. HINT [See Examples 1 and 2.] (Order your answe
icang [17]

Answer:

  • (-1, -32) absolute minimum
  • (0, 0) relative maximum
  • (2, -32) absolute minimum
  • (+∞, +∞) absolute maximum (or "no absolute maximum")

Step-by-step explanation:

There will be extremes at the ends of the domain interval, and at turning points where the first derivative is zero.

The derivative is ...

  h'(t) = 24t^2 -48t = 24t(t -2)

This has zeros at t=0 and t=2, so that is where extremes will be located.

We can determine relative and absolute extrema by evaluating the function at the interval ends and at the turning points.

  h(-1) = 8(-1)²(-1-3) = -32

  h(0) = 8(0)(0-3) = 0

  h(2) = 8(2²)(2 -3) = -32

  h(∞) = 8(∞)³ = ∞

The absolute minimum is -32, found at t=-1 and at t=2. The absolute maximum is ∞, found at t→∞. The relative maximum is 0, found at t=0.

The extrema are ...

  • (-1, -32) absolute minimum
  • (0, 0) relative maximum
  • (2, -32) absolute minimum
  • (+∞, +∞) absolute maximum

_____

Normally, we would not list (∞, ∞) as being an absolute maximum, because it is not a specific value at a specific point. Rather, we might say there is no absolute maximum.

5 0
3 years ago
2^7/2^3 basic exponent rules
Stells [14]

Answer:

Step-by-step explanation:

Since they have the same base as 2, by the rule of a^m / a^n = a^m-n.

So 2^7 / 2^3 = 2^ 7-3 = 2^4 = 16

6 0
2 years ago
Read 2 more answers
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