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3 years ago

12x+7<−11 OR 5x−8>40 solve for x

Mathematics

2 answers:

Flura [38]3 years ago

7 0

Answer:

x < -3/2 OR x > 9 3/5

Step-by-step explanation:

12x+7<−11 OR 5x−8>40

Solve the inequalities separately

12x+7<−11

Subtract 7 from each side

12x+7-7<−11 -7

12x < -18

Divide each side by 12

12x/12 < -18/12

x < -3/2

Then solve the other inequality

5x−8>40

Add 8 to each side

5x-8+8>40+8

5x>48

Divide each side by 5

5x/5 >48/5

x > 9 3/5

Put them back together with the OR in the middle

x < -3/2 OR x > 9 3/5

kipiarov [429]3 years ago

6 0

Answer:

x < -1.5 U x > 9.6

Step-by-step explanation:

12x + 7 < -11

x < -1.5

5x - 8 > 40

5x > 48

x > 9.6

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Answer:

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Step-by-step explanation:

We are told that the model of number of deers is;

y = 0.75x² + 10.5x + 1300

Where;

y represents number of deers

x represents number of years

Now, we want to find how many years it will take to get to 2000 deers.

Thus;

2000 = 0.75x² + 10.5x + 1300

0.75x² + 10.5x + 1300 - 2000 = 0

0.75x² + 10.5x - 700 = 0

Using quadratic formula, we have;

x = [-b ± √(b² - 4ac)]/2a

x = [-10.5 ± √(10.5²- (4*0.75*-700)]/(2*0.75)

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x = [-10.5 ± √2210.25]/(1.5)

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3 0

3 years ago

A. A 5in radius ball is in a

abruzzese [7]

Answer:

52.33%

Step-by-step explanation:

Since, all six sides of the box are touching the ball.

So, all the three dimensions of the box are equal.

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= diameter of the ball

= 2*radius of the ball

= 2*5

= 10 In.

$V_{ball} = \frac{4}{3} \pi r^3$

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$V_{ball} = \frac{4}{3} \times 3.14\times 125$

$V_{ball} = \frac{1,570}{3}\: In^3$

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Percentage of the space in the box occupied by the ball

$= \frac{V_{ball}}{V_{box} } \times 100$

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$= \frac{1570}{3000} \times 100$

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3 years ago

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3 years ago

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Jet001 [13]

Answer:

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3 years ago

State the converse, contrapositive, and inverse of each of these conditional statements a) If it snows tonight, then I will stay

marissa [1.9K]

Step-by-step explanation:

Consider the provided information.

For the condition statement $p \rightarrow q$ or equivalent "If p then q"

The rule for Converse is: Interchange the two statements.
The rule for Inverse is: Negative both statements.
The rule for Contrapositive is: Negative both statements and interchange them.

Part (A) If it snows tonight, then I will stay at home.

Here p is If it snows tonight, and q is I will stay at home.

Converse: If I will stay at home then it snows tonight.

$q \rightarrow p$

Inverse: If it doesn't snows tonight, then I will not stay at home.

$\sim p \rightarrow \sim q$

Contrapositive: If I will not stay at home then it doesn't snows tonight.

$\sim q \rightarrow \sim p$

Part (B) I go to the beach whenever it is a sunny summer day.

Here p is I go to the beach, and q is it is a sunny summer day.

Converse: It is a sunny summer day whenever I go to the beach.

$q \rightarrow p$

Inverse: I don't go to the beach whenever it is not a sunny summer day.

$\sim p \rightarrow \sim q$

Contrapositive: It is not a sunny summer day whenever I don't go to the beach.

$\sim q \rightarrow \sim p$

Part (C) When I stay up late, it is necessary that I sleep until noon.

P is I sleep until noon and q is I stay up late.

Converse: If I sleep until noon, then it is necessary that i stay up late.

$q \rightarrow p$

Inverse: When I don't stay up late, it is necessary that I don't sleep until noon.

$\sim p \rightarrow \sim q$

Contrapositive: If I don't sleep until noon, then it is not necessary that i stay up late.

$\sim q \rightarrow \sim p$

7 0

3 years ago