All categories

4 years ago

Which point is part of the solution of the inequality y ≤ |x + 2| − 3?

Mathematics

1 answer:

zhannawk [14.2K]4 years ago

6 0

You can try the points and see which satisfy the relation. I find it easier to consult a graph of the solution set and the offered points.

The appropriate choice is ...

... C: (1, 0) . . . . . . . on the boundary of the solution set.

You might be interested in

Unrjdjxbzbwnf c and f f

adelina 88 [10]

Answer:

Step-by-step explanation:

4 0

3 years ago

The sum of two numbers is

lara [203]

Answer:
The larger number is 60
The smaller number is 14

Explanation:
Assume that the larger number is x and that the smaller number id y.

We are given that:
1- The sum of the two numbers is 74. This means that:
x + y = 74
This equation can be rewritten as:
x = 74 - y .............> I
2- Four times the smaller number (4y) subtracted from the larger number (x), the result would be 4. This means that:
x - 4y = 4 ..............> II

Substitute with I in II and solve for y as follows:
x - 4y = 4
74 - y - 4y = 4
74 - 5y = 4
5y = 70
y = 70/5
y = 14

Substitute with the value of y in equation I to get x as follows:
x = 74 - y
x = 74 - 14
x = 60

Hope this helps :)

6 0

3 years ago

Read 2 more answers

HELP IM ABT TO FAIL :)

Delicious77 [7]

Answer:

15/36

Step-by-step explanation:

One hack that I used is I looked at all of the options in the first row and added the number one less 6 times

I basically did

5+(5-1)+(5-2)+(5-3)+(5-4)

5+4+3+2+1

7 0

4 years ago

Alexander placed four straws on his table.

Diano4ka-milaya [45]

Answer:D (straws 2 and 4)

Step-by-step explanation:

The answer is D becuase those two straws intersect at point and create 90° angle

6 0

3 years ago

The average of 16 consecutive positive integers is 30.5. What is the average of the first 8 integers of this set?

GREYUIT [131]

consecutive integers are 1 apart

16 of them can be represented as

x, x+1, x+2, x+3, x+4, x+5, x+6, x+7, x+8, x+9, x+10, x+11, x+12, x+13, x+14, x+15

the average is $\frac{x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15}{16}=30.5$

there might be some trick to get the average of the first 8 integers only

the average of the first 8 integers would be $\frac{x+x+1+x+2+x+3+x+4+x+5+x+6+x+7}{8}=?$

how can we get $\frac{x+x+1+x+2+x+3+x+4+x+5+x+6+x+7}{8}=?$ from $\frac{x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15}{16}=30.5$

first, let's match the denomenators (make them both 1 for ease)

multiply the first eqution by 8 to get

x+x+1+x+2+x+3+x+4+x+5+x+6+x+7=8?

multiply the 2nd equation by 16 to get

x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15=488

notice that we can try and force the 1st equation into the 2nd equation by adding some numbers

we can already do this:

x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15=488

[x+x+1+x+2+x+3+x+4+x+5+x+6+x+7]+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15=488

[8?]+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15=488

we can try to force another 8? into it

[8?]+x+7+1+x+7+2+x+7+3+x+7+4+x+7+5+x+7+6+x+7+7+x+7+8=488

[8?]+(x)+7+(1+x)+7+(2+x)+7+(3+x)+7+(4+x)+7+(5+x)+7+(6+x)+7+(7+x)+7+8=488

[8?]+(x)+7+(x+1)+7+(x+2)+7+(x+3)+7+(x+4)+7+(x+5)+7+(x+6)+7+(x+7)+7+8=488

group

[8?]+[x+x+1+x+2+x+3+x+4+x+5+x+6+x+7]+7+7+7+7+7+7+7+7+8=488

[8?]+[8?]+8(7)+8=488

16?+64=488

minus 64 from both sides

16?=424

divide both sides by 16

?=26.5

the average of the first 8 integers is 26.5

I'm not sure if there is a simpler way to do it or not

8 0

3 years ago