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3 years ago

For which of the following compound inequalities is there no solution?

Mathematics

2 answers:

gayaneshka [121]3 years ago

8 0

Answer:

3m - 12 > 30 and -6m >= 24

Step-by-step explanation:

A. 3m - 12 > 30 and -6m >= 24

3m > 42 and m < = -4

m > 14 and m < = -4

This has no solution

B. -6m >= 12 and m + 5 -18

cannot solve since missing inequality

C. -5m < 20 and 6m > -18

m > -4 and m > -3

solution m > -3

D. -4m - 10 <= -22 and 6m - 8 >= 22

-4m < = -12 and 6m > = 30

m > = 3 and m > =5

m > = 5

meriva3 years ago

4 0

Answer:

Step-by-step explanation:

$3m - 12 > 30 \text{ and } -6m\geq 24$

$\boxed{3m - 12 > 30 \wedge -6m\geq 24}$

$m>14 \wedge m\leq -4$

There's no solution. It is the first one already. There is no number that is both greater than 14 and less than or equal to -4. That is no solution because there's no $m$ that satisfy the compound inequality.

<u>Note: the signal change because we divided by negative number.</u>

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AnnyKZ [126]

Answer:

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Step-by-step explanation:

1: if you see something to the power you will always think of repeated multiplication off the bat. so you just multiply the coefficient by the amount of the exponent.

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3 years ago

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ziro4ka [17]

Answer:

$252.5

Step-by-step explanation:

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6 0

2 years ago

Vitek1552 [10]

To solve this we are going to use the future value of annuity due formula: $FV=(1+ \frac{r}{n} )*P[ \frac{(1+ \frac{r}{n})^{kt}-1 }{ \frac{r}{n} } ]$
where
$FV$ is the future value
$P$ is the periodic deposit
$r$ is the interest rate in decimal form
$n$ is the number of times the interest is compounded per year
$k$ is the number of deposits per year

We know for our problem that $P=420$ and $t=15$ . To convert the interest rate to decimal form, we are going to divide the rate by 100%: $r= \frac{10}{100} =0.1$ . Since Ruben makes the deposits every 6 months, $k=2$ . The interest is compounded semiannually, so 2 times per year; therefore, $k=2$ .
Lets replace the values in our formula:

$FV=(1+ \frac{r}{n} )*P[ \frac{(1+ \frac{r}{n})^{kt}-1 }{ \frac{r}{n} } ]$
$FV=(1+ \frac{0.1}{2} )*420[ \frac{(1+ \frac{0.1}{2})^{(2)(15)}-1 }{ \frac{01}{2} } ]$
$FV=29299.53$

We can conclude that the correct answer is <span>$29,299.53</span>

8 0

3 years ago

Which of the following is the sum of all the odd integers between 5 and 102?

STALIN [3.7K]

Answer:

2601

Step-by-step explanation:

The integers between 5 and 102 are

7, 9, 11, 13, 15, 17,19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101

The sum of the odd intergers between 2 and 105 can be calculated as follows

= (N + 1/2)^2

= (101 + 1 /2) ^2

= (102/2)^2

= 51^2

= 2601

Hence the sum of the odd intergers between 5 and 102 is 2601

3 0

3 years ago

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daser333 [38]

Answer: