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8_murik_8 [283]
3 years ago
15

Did I get all of these right? feedback from experts please!

Mathematics
1 answer:
Troyanec [42]3 years ago
8 0

Answer:

ucf8riucfuvigruvufgv

Step-by-step explanation:

u7dfi8t57dch

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It’s a simple algebra question please answer the picture
Shtirlitz [24]

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)

4 0
3 years ago
I'm studying for a quiz but I don't know how to do these two questions! Can someone explain them to me?
Lemur [1.5K]

Answer:

yes mate, what r ur questions??

☯ ♡ ♛ ┈ ⛧ ┈ ◍ °◇◆◇ °❣❥➻° A꙰ N꙰꙰ N꙰꙰ °➻❥❣️° ◆◇◆° ◍ ┈ ⛧ ┈ ♛ ♡ ☯

5 0
3 years ago
Read 2 more answers
Heyy!! Can someone help me please!!
abruzzese [7]

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

5 0
2 years ago
Read 2 more answers
What is the equation, in point-slope form, of the line that is perpendicular to the given line and passes through the
Svetradugi [14.3K]

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

\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\underline{1}\implies \cfrac{\underline{1}}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{\underline{1}}}\qquad \stackrel{negative~reciprocal}{-\cfrac{1}{\underline{1}}\implies -1}}

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

\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{5})~\hspace{10em} \stackrel{slope}{m}\implies -1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{5}=\stackrel{m}{-1}(x-\stackrel{x_1}{2}) \\\\\\ y-5=-x+2\implies y=-x+7

8 0
2 years ago
Read 2 more answers
5g + 4(-5+3g) = 1 - g
Korvikt [17]
5g + 4(-5 + 3g) = 1 - g 
5g - 20 + 12g = 1 - g 
 17g - 20 = 1 - g 
   \frac{18g}{18} =   \frac{21}{18}
    there can be to answer 
      g = \frac{21}{18} or 
      you can simplify \frac{21/3}{18/3} =  \frac{7}{6} or 1 \frac{1}{6}
 
 
8 0
2 years ago
Read 2 more answers
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