In the diagram below, de is parallel to xy. What is the value of y?

I have no idea what the answers are

lara [203]

Hello from MrBillDoesMath!

Answer:

Top line: y = (2/3)x + 2

Bottom line: y = (2/3)x -1

Discussion:

The graph provided is hard to read but I did the best I could.

The top line appears to pass through the points (0,2) and (-3,0)

For this line

m = change y /change x = (0-2)/(-3-0) = -2/-3 = +2/3. So

y = mx + b => y = (2/3) x+ b. As the line passes through (0,2) set x = 0, y= 2 in y = (2/3)x + b =>

2 = (2/3) 0 + b => b = 2

Therefore y = (2/3)x + 2

The bottom line appears to pass through the points (0,-1) and (3,1)

For this line

m = change y /change x = (1-(-1)) /(3-0) = +2/-3. So

y = mx + b => y = (2/3) x+ b. As the line passes through (0,-1) set x = 0, y= -1 in y = (2/3)x + b =>

-1 = (2/3) 0 + b => b = -1

Therefore y = (2/3)x + -1

Thank you,

MrB

5 0

2 years ago

10 points! I will thanks, rate, and give best answer!!

vfiekz [6]

Answer: The fourth term is -102

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Explanation:

The term after the nth term is generated by this rule $a_{n+1} = -4(a_n) + 2$ which means that we first
Step 1) multiply the nth term ( $a_n$ ) by -4
Step 2) Add the result of step 1 to the value 2 to get the next term in the sequence

Let's follow those steps above to generate the first four terms

The first term is $a_1 = 2$ . In short, the first term is 2

The second term is...
$a_{n+1} = -4(a_n) + 2$
$a_{1+1} = -4(a_1) + 2$
$a_{2} = -4(2) + 2$
$a_{2} = -8 + 2$
$a_{2} = -6$
So the second term is -6

The third term is...
$a_{n+1} = -4(a_n) + 2$
$a_{2+1} = -4(a_2) + 2$
$a_{3} = -4(-6) + 2$
$a_{3} = 24 + 2$
$a_{3} = 26$
The third term is 26

Finally, the fourth term is...
$a_{n+1} = -4(a_n) + 2$
$a_{3+1} = -4(a_3) + 2$
$a_{4} = -4(26) + 2$
$a_{4} = -104 + 2$
$a_{4} = -102$
The fourth term is -102.

4 0

3 years ago

In ΔABC, c = 5.4, a = 3.3, and . What are the possible approximate lengths of b? Use the law of sines to find the answer. 2.0 un

Aloiza [94]

Correct answer is C) 2.3 units and 7.8 units

5 0

2 years ago

Given the acceleration, initial velocity, and initial position of a body moving along a coordinate line at time t, find the body

nika2105 [10]

Answer:

The position at time t is $s(t) = 8t^2 - 14t -8$