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Arte-miy333 [17]
4 years ago
6

Which counterexample proves this statement is FALSE?

Mathematics
1 answer:
Nina [5.8K]4 years ago
8 0

Answer:

The answer is C.

Step-by-step explanation:


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Find Mx, My, and (x, y) for the lamina of uniform density rho bounded by the graphs of the equations. y = x2/3, y = 0, x = 1
erik [133]

Answer:

\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8},  \dfrac{5}{14})}

Step-by-step explanation:

Given that:

y =  x^{2/3} at y = 0 , x = 1

Then:

Area = \int^{1}_{0} x^{2/3} \ dx

Area = \begin {bmatrix} \dfrac{3}{5}x^{5/3} \end {bmatrix} ^1_0

Area = \dfrac{3}{5}

Then:

\overline x = \dfrac{1}{A} \int^b_a x (f(x) -g(x) ) \ dx

\overline x = \dfrac{5}{3} \int^1_0 x (x^{2/3} -0 ) \ dx

\overline x = \dfrac{5}{3} \int^1_0 x^{5/3} \ dx

\overline x = \dfrac{5}{3} \ [\dfrac{3}{8}x^{8/3}]^1_0

\overline x = \dfrac{5}{3} \times \dfrac{3}{8}

\overline x = \dfrac{5}{8}

Similarly;

\overline y = \dfrac{1}{A} \int^b_a \dfrac{1}{2} \begin{bmatrix} (f(x)^)2 - (g(x))^2 \end {bmatrix}  \ dx

\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (f(x^{2/3})^2 -0 \end {bmatrix}  \ dx

\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (x^{4/3} ) \end {bmatrix}  \ dx

\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{1}{2}  (x^{7/3} ) \times \dfrac{3}{7} \end {bmatrix} ^1_0

\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{3}{14}  (x^{7/3} ) \end {bmatrix} ^1_0

\overline y = (\dfrac{5}{3} \times \dfrac{3}{14} )

\overline y = \dfrac{5}{14}

Thus; \mathbf{(\overline x , \overline y ) = (\dfrac{5}{8},  \dfrac{5}{14})}

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3 years ago
Which probability indicates that an event will likely occur? A- 0 B- 1/4 C- 1/2 D- 3/4
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The answer is c- 1/2<span />
6 0
3 years ago
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Which of the following values is a solution to the inequality? 12 – 5x &gt; -8
olya-2409 [2.1K]

Answer:

x < 4

Step-by-step explanation:

12 – 5x > -8

Subtract 12 from each side

12-12 – 5x > -8-12

-5x> -20

Divide each side by -5, remembering to flip the inequality

-5x/-5 < -20/-5

x < 4

5 0
3 years ago
explain how you can tell whether a group of ratios represents a proportional relationship. PLEASE HELP
zubka84 [21]
Ratios are proportional if they represent the same relationship. One way to see if two ratios are proportional is to write them as fractions and then reduce them. If the reduced fractions are the same, your ratios are proportional.

hope this helps :)
6 0
3 years ago
A number cube with faces labeled from 1 to 6 will be rolled once.
tresset_1 [31]

Question:

A number cube with faces labeled from 1 to 6 will be rolled once.  The number rolled will be recorded as the outcome.

Give the sample space describing all possible outcomes.  Then give all of the outcomes for the event of rolling a number greater than 2.

If there is more than one element in the set, separate them with commas.

Answer:

S = \{1,2,3,4,5,6\}

Greater2 = \{3,4,5,6\}

Step-by-step explanation:

Given

A roll of a 6 sided number cube

Solving (a): The sample space

This implies that we list out all number on the number cube.

So:

S = \{1,2,3,4,5,6\}

Solving (b): Outcomes greater than 2

This implies that we list out all number on the number cube greater than 2 i.e. 3 to 6.

So:

Greater2 = \{3,4,5,6\}

3 0
3 years ago
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