Answer:
JL = 62 (units)
Step-by-step explanation:
OK, Here we can use the part whole postulates theorem to assist us:
JL consists of parts JK + KL such that, through substitution,
5x + 2 = 27 + ( 3x - 1 )
Here we can solve for x to determine 5x + 2, and thus the value of JL:
5x + 2 = 27 + 3x - 1
2x + 2 = 27 - 1
2x + 2 = 26
2x = 24
x = 12
Substitute the value of x for 5x + 2, to get JL:
JL = 5 (12) + 2
JL = 60 + 2
JL = 62 (units)
Answer:
yes mate, what r ur questions??
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Answer:
7x + 14
Step-by-step explanation:
the first thing to do is expand the parentheses/brackets.
3(5x + 2) -2(4x - 4) will be
3(5x) + 3(2) -2(4x) -2(-4)
= 15x + 6 - 8x + 8
collect like terms
15x - 8x + 6 + 8 = 7x + 14
the answer is 7x + 14
keeping in mind that perpendicular lines have negative reciprocal slopes, hmmmm what's the slope of that line above anyway,
![\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{(-1)}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{1}}}\implies \cfrac{2+1}{3}\implies 1 \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B-1%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B4%7D~%2C~%5Cstackrel%7By_2%7D%7B2%7D%29%20~%5Chfill%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%20%7B%5Cstackrel%7By_2%7D%7B2%7D-%5Cstackrel%7By1%7D%7B%28-1%29%7D%7D%7D%7B%5Cunderset%7Brun%7D%20%7B%5Cunderset%7Bx_2%7D%7B4%7D-%5Cunderset%7Bx_1%7D%7B1%7D%7D%7D%5Cimplies%20%5Ccfrac%7B2%2B1%7D%7B3%7D%5Cimplies%201%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

so we're really looking for the equation of a line whose slope is -1 and runs through (2,5)
