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stealth61 [152]
2 years ago
10

Find the value of x. Please I really need help

Mathematics
2 answers:
jonny [76]2 years ago
7 0

Answer:

x = 16

Step-by-step explanation:

180=(5x-8)+(7x-4)

180=12x-12

192/12=x

x=16

Serggg [28]2 years ago
7 0

Answer:

I think 12x is that an option

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Can someone help me with this ? Thanks
olasank [31]

Answer:

Type of angle - Obtuse

Key information - 180° - 82° = 98° (missing angle)

Equation - (5y+12)=98 y=17.2

Step-by-step explanation:

Let's solve for the missing angle first! 180° - 82° = 98°. 180 represents the total among the angles so we use 180 degrees.

then use the statement to solve for y and make it equal to 98°

(5y+12)=98

(5y-12)  = -12          subtract 12 from both sides

(5y)           =  86        Now solve for y by dividing 86 ÷ 5

y = 17.2                      

5 0
3 years ago
Read 2 more answers
"Lito had some marbles. He gave ½ of his marbles plus 1 to Felix, ½ of the remaining marbles plus 1 to Ces, and ½ of the last re
Nuetrik [128]

let

x-------> total amount of marbles at the beginning


we know that

1) He gave ½ of his marbles plus 1 to Felix

(1/2)*x+1=(x+2)/2 (Felix's marbles)

remaining=x-[(x+2)/2]-----> [2x-x-2]/2------> (x-2)/2


2)½ of the remaining marbles plus 1 to Ces

(1/2)*[(x-2)/2]+1=[(x-2)/4]+1-----> (x+2)/4 (Ce's marbles)

remaining=[(x-2)/2]-[(x+2)/4]-----> [2x-4-x-2]/4------> (x-6)/4


3) ½ of the last remaining marbles plus 1 to Pedro

(1/2)*[ (x-6)/4]+1=[(x-6)/8)+1------> (x+2)/8 (Pedro's marbles)

remaining=[(x-6)/4]-[(x+2)/8]------> [2x-12-x-2]/8-----> (x-14)/8


4)If Lito had 1 marble left for himself

so

the last remaining is equal to 1

(x-14)/8=1-----> x-14=8------> x=22 marbles


Verify

(x+2)/2 (Felix's marbles)------> (22+2)/2=12

(x+2)/4 (Ce's marbles)------> (22+2)/4=6

(x+2)/8 (Pedro's marbles)---> (22+2)/8=3

Lito's marbles------------------> 1

total=12+6+3+1=22--------> is ok


therefore


the answer is

the total amount of marbles at the beginning was 22

4 0
3 years ago
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The volume is 10 cubic meters  is 6 cenemeters
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