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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.

Note: the signal change because we divided by negative number.

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Answer:

x=-20

Step-by-step explanation:

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3 years ago

If f(x) = 7x − 1, what does f(12) represent?

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The coordinates of polygon ABCD are A(-4,-1), B(-2,3), C(2,2), and D(4,-3). Use these coordinates to complete the sentences belo

PolarNik [594]

Answer:

The perimeter of polygon ABCD, to the nearest thousandth units is

22.227 units

The perimeter of polygon ABCD', to the nearest thousandth, would be 20.980 units

The area of polygon ABCD' is 19.5 units²

Step-by-step explanation:

The coordinates forming the polygon are

A (-4,-1), B(-2,3), C(2,2), and D(4,-3)

The perimeter then is given by the sum of the length of the sides as follows;

Length of line between two X and Y points distance, xi and yj apart is

length XY = $\sqrt{x^2+y^2}$

Therefore, the length between points AB is

length AB = $\sqrt{(-4 - (-2))^2+(-1-3)^2}$

= $\sqrt{(-4 +2)^2+(-4)^2} = \sqrt{-2^2+-4^2} = \sqrt{20}$ = 4.472 units

Similarly, length BC is given by

Length BC = $\sqrt{(-4)^2+1^2} = \sqrt{17}$ = 4.123 units

Length CD = $\sqrt{(-2)^2+5^2} = \sqrt{29}$ = 5.385 units

Length DA = $\sqrt{(8)^2+(-2)^2} = \sqrt{68}$ = 8.246 units

The perimeter is equal to;

length AB + Length BC + Length CD + Length DA

= 4.472 + 4.123 + 5.385 + 8.246 = 22.227 units

If the point D is moved up 2 units and left 1 unit we have

D' = x-1, y+2 where x and y are the coordinates of point D

D(4, -3) → D'((4-1), (-3+2)) = D'(3, -1)

The length D'A = $\sqrt{(7)^2+(0)^2} = \sqrt{49}$ =7 units

The perimeter of polygon ABCD'

length AB + Length BC + Length CD + Length D'A

= 4.472 + 4.123 + 5.385 + 7 = 20.980 units.

The area is given by the determinant of the 3 by 3 matrix using Cramer's Rule as follows

$S_{\bigtriangleup}$ = $(\frac{1}{2})|x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2|$

= $(\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|$

Triangle ABC we have

A(-4,-1)

B(-2,3)

C(2,2)

$S_{{\bigtriangleup}ABC}$ = $(\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|$

$=(\frac{1}{2})|(-4)(3-2)+(-2)(2-(-1)) +2((-1)-3)|$

= $=(\frac{1}{2})|-4-6 -8|= 9 units^2$

Triangle ACD'

A(-4,-1)

C(2,2)

D'(3, -1)

$S_{{\bigtriangleup}ACD'}$ = $(\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|$

$=(\frac{1}{2})|(-4)(2+1)+(2)(-1+1) +3((-1)-2)| = \frac{21}{2}$ = 10.5 units²

Area of polygon = $S_{{\bigtriangleup}ABCD'}$ = $S_{{\bigtriangleup}ABC}$ + $S_{{\bigtriangleup}ACD'}$ = (9 + 10.5) units²

Area of polygon ABCD' = 19.5 units².

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3 years ago

You want to distribute 7 candies to 4 kids. If every kid must receive at least one candy, in how many ways can you do this?

puteri [66]

Answer:

Step-by-step explanation:

lets begin to set up this question. i personally find questions like these easiest when i begin to set up ideas to help my brain process it better, for example:

person a

person b

person c

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we know that each (person) must receive at least 1 candy. we also know that we have 7 candies. therefore if we want do draw this out :

person a- 1

person b- 1

person c- 1

person d -1

notice that we now have given away 4 of the 7 candies. now we have 3 candies left. we can simply:

person a- 2

person b-2

person c-2

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but, keep in mind that there are still many more ways that we can distribute the candies to other kids. the question we are asked is how may ways can we do this. since i have already illustrated the question, you can either learn how to put this into an equation, or experiment how many variations there are. the answer, either way is 15.

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3 years ago

A triangle has two sides of length 3 and 3. What value could the length of thethird side be? Check all that apply.A. 6B. 12C. 5D

Bumek [7]

To find out the length of the third segments of the triangle, we can use this formula.

$\begin{gathered} a^2=b^2+c^2 \\ a^2=3^2+3^2 \\ a^2=9+9 \\ a^2=18 \\ a=4.25 \end{gathered}$

The, third length is very close to the 4. thus, the option (E) is correct.

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10 months ago