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3 years ago

Which expression is equivalent to (18x + 9k) + (2x + 6k)?

Mathematics

2 answers:

uysha [10]3 years ago

7 0

Answer:

Step-by-step explanation:

the person above me should have brainliest

Marrrta [24]3 years ago

5 0

Answer:

B. 20x + 15

Step-by-step explanation:

Get rid of the parenthesis and combine like terms: 18x + 2x and 9k + 6k

You get: 20x + 15.

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Solving Linear Systems: Using Substitution <br> x=3y-4<br> 3y - 4x = 6<br> someone pls help

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Answer:

Point form : (- 2/3, 10/9)

Equation form : x=- 2/3, y=10/9

Step-by-step explanation: Solve for the first variable in one of the equations, then substitute the result into the other equation.

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How do you turn a inproper mixed fraction proper

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Five computer program modules are ranked as M1, M2, M3, M4, and M5 according to the ascending order of effort required to debug

gulaghasi [49]

Answer:

Follows are the solution to this question:

Step-by-step explanation:

Technician selects three out of 5 systems

In C(5,3)=10ways, this can be achieved

In part a:

Space sample chooses 3 of a 5 systems

$(M_1, \ M_2,\ M_3),$ $(M_1,M_2,M_4) \ (M_1,M_2,M_5) \ (M_1,M_3,M_4) \ (M_1,M_3,M_5),$ $(M_1,M_4,M_5) \ (M_2,M_3,M_4)\ (M_2,M_3,M_5) \ (M_2,M_4,M_5),(M_3,M_4,M_5)}$

In point b:

A =MODULE WHICH INCLUDE M1 minimal amount of effort

Outcomes probable =

$(M_1,M_2,M_3),\ (M_1,M_2,M_4) \ (M_1,M_2,M_5)\ (M_1,M_3,M_4)\\\\(M_1,M_3,M_5),\ (M_1,M_4,M)5)\ =\ 6$

$\to p(A)=\frac{6}{10}\\\\$

$=0.6$

In point c:

B = highest effort that is $M_5$

Potential result=

$(M_1,M_2,M_5) \ (M_1,M_3,M_5) \ (M_2,M_3,M_5)\(M_2,M_4,M_5) \\ (M_2,M_4,M_5), \ (M_3,M_4,M_5) \ =\ 6 \\\\$

$\to B= \frac{6}{10} \\\\$

$=0.6$

$\to P(B)=10$

In point d:

$\to \ A \ intersection \ B=(M_1,M_2,M_5), \ (M_1,M_3,M_5) \ ,(M_1,M_4,M_5)$

$\to A (A \ intersection \ B) = \frac{3}{10} \\\\\ \ \ \ \ \$

$=0.3$

In point e:

$\to (A \cup B) = (M_1,M_2,M_3),\ (M_1,M_2,M_4)\ (M_1,M_2,M_5)(M_1,M_3,M_4)\ (M_1,M_3,M_5), \\ (M_1,M_4,M_5)\ (M_2,M_3,M_5) \ (M_2,M_4,M_5),(M_3,M_4,M_5) \ = \ 9$ $\to P(A \cap B)=\frac{9}{10}$