Ask question

Ask question

All categories

4 years ago

12

Which of the following is a solution of the system

1 answer:

Aleks [24]4 years ago

6 0

Answer:

Answer B

Step-by-step explanation:

1. (1, -4) is NOT a solution because (1) - 2(-4) < 4 is false.

2. (-8, 2): -8 - 2(2) < 4 is true, and in simplified form is -12 < 4 (true) We must check whether this point (-8, 2) ALSO satisfies y > -2x - 5:

2 > -2(-8) - 5. Yes, this is true, so the given point satisfies both inequalities in Answer B.

You might be interested in

2k^2-5k-18=0 what is both of the values of k? plz help!!! 15 pts!!!

ddd [48]

To factor quadratic equations of the form ax^2+bx+c=y, you must find two values, j and k, which satisfy two conditions.

jk=ac and j+k=b

The you replace the single linear term bx with jx and kx. Finally then you factor the first pair of terms and the second pair of terms. In this problem...

2k^2-5k-18=0

2k^2+4k-9k-18=0

2k(k+2)-9(k+2)=0

(2k-9)(k+2)=0

so k=-2 and 9/2

k=(-2, 4.5)

5 0

3 years ago

Which of the following is most likely the next step in the series?<br>

Andrej [43]

Answer:

B

Step-by-step explanation:

They are increasing by 1 vertically. Hope this helps!! :)

5 0

3 years ago

Read 2 more answers

What is 153/4 as a whole number

maw [93]

$\frac{4}{3} = \frac{51}{n} \\ = > \frac{3}{4} = \frac{n}{51 } \\ = > n = \frac{3}{4} \times 51 \\ = > n = \frac{153}{4} \\ = > n = 38 \frac{1}{4}$

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0

2 years ago

Find the x-intercept and the y-intercept of 2x + 6y =24

viktelen [127]

Answer:

x-intercept = 12, y-intercept = 4

Step-by-step explanation:

To find the x and y intercept, substitute 0 in for both values.

<u>Substitute 0 into x:</u>

2(0) + 6y = 24

6y = 24

<u>Divide each side by 6:</u>

y = 4

<u>Substitute 0 into y:</u>

2x + 6(0) = 24

2x = 24

<u>Divide each side by 2:</u>

x = 12

4 0

2 years ago

The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop

Anettt [7]

Answer:

a. $\frac{1}{15}$

b. $\frac{2}{5}$

c. $\frac{14}{15}$

d. $\frac{8}{15}$

Step-by-step explanation:

Given that there are two laptop machines and four desktop machines.

On a day, 2 computers to be set up.

To find:

a. probability that both selected setups are for laptop computers?

b. probability that both selected setups are desktop machines?

c. probability that at least one selected setup is for a desktop computer?

d. probability that at least one computer of each type is chosen for setup?

Solution:

Formula for probability of an event E can be observed as:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

a. Favorable cases for Both the laptops to be selected = $_2C_2$ = 1

Total number of cases = 15

Required probability is $\frac{1}{15}$ .

b. Favorable cases for both the desktop machines selected = $_4C_2=6$

Total number of cases = 15

Required probability is $\frac{6}{15} = \frac{2}{5}$ .

c. At least one desktop:

Two cases:

1. 1 desktop and 1 laptop:

Favorable cases = $_2C_1\times _4C_1 = 8$

2. Both desktop:

Favorable cases = $_4C_2=6$

Total number of favorable cases = 8 + 6 = 14

Required probability is $\frac{14}{15}$ .

d. 1 desktop and 1 laptop:

Favorable cases = $_2C_1\times _4C_1 = 8$

Total number of cases = 15

Required probability is $\frac{8}{15}$ .

8 0

4 years ago