Answer:
b
Step-by-step explanation:
Given K is the midpoint of JL, then
JK = KL ← substitute values
6x = 3x + 3 ( subtract 3x from both sides )
3x = 3 ( divide both sides by 3 )
x = 1
Hence
JK = 6x = 6 × 1 = 6
KL = 3x + 3 = (3 × 1) + 3 = 3 + 3 = 6
JL = 6 + 6 = 12
Answer:
125x³ − 150x²y + 60xy² − 8y³
Step-by-step explanation:
(5x − 2y)³
₃C₀ (5x)³ (-2y)⁰ + ₃C₁ (5x)² (-2y)¹ + ₃C₂ (5x)¹ (-2y)² + ₃C₃ (5x)⁰ (-2y)³
(1) (125x³) (1) + (3) (25x²) (-2y) + (3) (5x) (4y²) + (1) (1) (-8y³)
125x³ − 150x²y + 60xy² − 8y³
Answer:
a = 6/5
b = 7/8
c = 1/5
Step-by-step explanation:
We want to make two of these equations be in terms of the same variable, so let's solve the first and third in terms of b.
5a-24b=-15 -> -3 + 24b/5 = a
48b+35c = 49 -> 49/35 - 48b/35 = 7/5 - 48b/35 = c
Now we can replace the a and c in the second equation and solve for b.
10a+45c=21
10(-3+24b/5)+45(7/5 - 48b/35) = 21
-30 + 48b + 63 - 432b/7 = 21 -> -12 = -96b/7 -> b=7/8
Now we can plug b back into the other two to solve for a and c. I will leave that to you unless you would like the steps there.
Answer:
Table 4 represents the same linear expression as y = 3 x - 2.
Step-by-step explanation:
Here, the given expression is y = 3 x -2
So now check the any random pair of each table by putting in the given equation.
<u>TABLE 1 : (2,-5)</u>
y = 3x -2 ⇒ -5 = 3(2) -2
or, -5 = -4 , NOT POSSIBLE
<u>TABLE 2 : (0,3)</u>
y = 3x -2 ⇒ 3 = 3(0) -2
or, 3 = -2 , NOT POSSIBLE
<u>TABLE 3 : (1,2)</u>
y = 3x -2 ⇒ 2 = 3(1) -2
or, 2 = 1 , NOT POSSIBLE
<u>TABLE 4 : (1,1)</u>
y = 3x -2 ⇒ 1 = 3(1) -2
or, 1 = 1 , POSSIBLE
checking for (2,4)
4 = 3(2) - 2 ⇒4 = 4 POSSIBLE
Here, table 4 satisfies the given points in the expression y = 3x -2
Hence, it represents the same linear expression .