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Svetlanka [38]
2 years ago
14

Read the paragraph. [1] The 1992 Olympic Games took place in Barcelona, Spain. [2] This marked the first year that professional

basketball players were permitted to represent their countries in the games. [3] Previously, only amateur players were allowed to participate such as college athletes. [4] The US squad was called “The Dream Team” because it consisted of some of the best players in the history of the game, including Michael Jordan, Magic Johnson, and Larry Bird. Which sentence should be revised to improve this paragraph’s sentence fluency?Which are correct representations of the inequality –3(2x – 5) < 5(2 – x)? Select two options. x < 5 –6x – 5 < 10 – x –6x + 15 < 10 – 5x A number line from negative 3 to 3 in increments of 1. An open circle is at 5 and a bold line starts at 5 and is pointing to the right. A number line from negative 3 to 3 in increments of 1. An open circle is at negative 5 and a bold line starts at negative 5 and is pointing to the left.
Mathematics
1 answer:
tatuchka [14]2 years ago
4 0

The sentence that needs to be revised to improve the paragraph's sentence fluency is: sentence 3.

The two correct representations for the inequality can be -6x + 15 < 10 - 5x OR x < 5

In a grammatical context, the correct usage of punctuation marks and active voice is required and needed to be taken into consideration by the writer for effective communication to the reader.

The correct form of the sentence in sentence 3 should be:

  • "Previously, only amateur players, like college athletes, were allowed to participate."

The inequality given is -3(2x - 5) < 5(2 - x). Using the simple algebraic method to solve the given inequality, we have:

= -3(2x - 5) < 5(2 - x)

= -6x + 15 < 10 - 5x

By collecting the like terms, we have:

= 15 - 10 < -5x + 6x

= x < 5

Therefore, the two correct representations for the inequality can be -6x + 15 < 10 - 5x  OR  x < 5

NOTE: The third question is a repetition of an incomplete question.

Learn more about sentences here:

brainly.com/question/4955765

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Which of the following is the solution to the inequality below ? -6/5(4-x)&lt;_-4(X+1/2)
LuckyWell [14K]

Answer:

The answer is the option D

x\leq 7/13

Step-by-step explanation:

we have

-\frac{6}{5}(4-x)\leq -4(x+\frac{1}{2})

Multiply by -5 both sides

6(4-x)\geq 20(x+0.5)\\24-6x\geq20x+10\\24-10\geq 20x+6x\\14\geq 26x

Rewrite

26x\leq 14\\x\leq 14/26\\ x\leq 7/13


7 0
2 years ago
Use the law of cosines to find the value of 2*3*4 cos theta.
daser333 [38]
The law of cosines :

A^2 = B^2 + C^2 - 2BC cos A

2^2 = 3^2 + 4^2 -  2 . 3. 4 cos theta

= 21

hope this helps
3 0
3 years ago
Read 2 more answers
Evaluate the formula V=BH/3 for B=4cm^2 and h=6 cm
Tom [10]
V=BH/3
V=(4)(6)/3
V=(24)/3
V=8
6 0
3 years ago
Read 2 more answers
9 (1/3)^n-1 what is the tenth term
NARA [144]

Answer:

a_{10}=\frac{1}{2187}

Step-by-step explanation:

Simply plug in 10 for <em>n</em>

9(1/3)¹⁰⁻¹

9(1/3)⁹

9(1/19683)

9/19683

a_{10}=\frac{1}{2187}

7 0
3 years ago
Find the point on the parabola y^2 = 4x that is closest to the point (2, 8).
guapka [62]

Answer:

(4, 4)

Step-by-step explanation:

There are a couple of ways to go at this:

  1. Write an expression for the distance from a point on the parabola to the given point, then differentiate that and set the derivative to zero.
  2. Find the equation of a normal line to the parabola that goes through the given point.

1. The distance formula tells us for some point (x, y) on the parabola, the distance d satisfies ...

... d² = (x -2)² +(y -8)² . . . . . . . the y in this equation is a function of x

Differentiating with respect to x and setting dd/dx=0, we have ...

... 2d(dd/dx) = 0 = 2(x -2) +2(y -8)(dy/dx)

We can factor 2 from this to get

... 0 = x -2 +(y -8)(dy/dx)

Differentiating the parabola's equation, we find ...

... 2y(dy/dx) = 4

... dy/dx = 2/y

Substituting for x (=y²/4) and dy/dx into our derivative equation above, we get

... 0 = y²/4 -2 +(y -8)(2/y) = y²/4 -16/y

... 64 = y³ . . . . . . multiply by 4y, add 64

... 4 = y . . . . . . . . cube root

... y²/4 = 16/4 = x = 4

_____

2. The derivative above tells us the slope at point (x, y) on the parabola is ...

... dy/dx = 2/y

Then the slope of the normal line at that point is ...

... -1/(dy/dx) = -y/2

The normal line through the point (2, 8) will have equation (in point-slope form) ...

... y - 8 = (-y/2)(x -2)

Substituting for x using the equation of the parabola, we get

... y - 8 = (-y/2)(y²/4 -2)

Multiplying by 8 gives ...

... 8y -64 = -y³ +8y

... y³ = 64 . . . . subtract 8y, multiply by -1

... y = 4 . . . . . . cube root

... x = y²/4 = 4

The point on the parabola that is closest to the point (2, 8) is (4, 4).

4 0
3 years ago
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