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Anestetic [448]
2 years ago
13

15

Mathematics
1 answer:
IRISSAK [1]2 years ago
5 0

Answer:

decrease because the ratio number is Less so multiply by 156 to each ratio to find your relationship between. hope this helps.

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Riverside Elementary School is holding a school-wide election to choose a school color. 5/8 of the voters were for blue 5/9 of t
padilas [110]

Answer:

There were 180 voters for blue color.

Step-by-step explanation:

Let the total number of voters be 'x'.

Given:

Number of Voters for blue color = \frac{5}{8}x

Number of Voters for green color = \frac{5}{9}(x-\frac58x)=\frac59x(1-\frac58)

Now we will use LCM to make the denominator common we get;

Number of Voters for green color = \frac59x(\frac88-\frac58)=\frac59x(\frac{8-5}{9})=\frac59x(\frac38)=\frac{15x}{72}

Number of Voters for red color = 48

We need to find the number of voters for blue color.

Solution:

Now we can say that;

total number of voters is equal to sum of Number of Voters for blue color, Number of Voters for green color and Number of Voters for red color.

framing in equation form we get;

x=\frac58x+\frac{15x}{72}+48

Combining like terms we get;

x-\frac58x-\frac{15x}{72} = 48

Now we will make denominators common using LCM we get;

\frac{72x}{72}-\frac{5x\times9}{8\times9}-\frac{15x\times1}{72\times 1} = 48\\\\\frac{72x}{72}-\frac{45x}{72}-\frac{15x}{72} = 48

Now denominators are common so we will solve the numerators we get;

\frac{72x-45x-15x}{72}=48\\\\\frac{12x}{72}=48\\\\\frac{x}{6}=48

Now multiplying both side by 6 we get;

\frac{1}{6}x\times6=48\times6\\\\x = 288

Number of voters for blue color = \frac{5}{8}x=\frac{5}{8}\times 288= 180

Hence There were 180 voters for blue color.

4 0
3 years ago
Solve for x4:=(-8)+3x
Bingel [31]
I added 8 on both sides to get 12=3x then I divided by 3 on both sides to get x=4
7 0
3 years ago
What are the greatest common factors of 51 and 85
rusak2 [61]
Greatest\ common\ factor\ is\ the\ highest\ number\ by\ which\ 51\ and\ 85\\
can\ be\ divided\\\\ 
51:51\\\\
85:5\\
17:17\\\\GCF=1\\\\Greatest\ common\ factor\ of\ 51\ and\ 85\ is\ equal\ to\ 1.

6 0
3 years ago
Prove that the diagonals of a parallelogram bisect each other​
Nady [450]

Answer:

[ See the attached picture ]

The diagonals of a parallelogram bisect each other.

✧ Given : ABCD is a parallelogram. Diagonals AC and BD intersect at O.

✺ To prove : AC and BD bisect each other at O , i.e AO = OC and BO = OD.

Proof :\begin{array}{ |c| c |  c |  } \hline \tt{SN}& \tt{STATEMENTS} & \tt{REASONS}\\ \hline 1& \sf{In  \: \triangle ^{s}  \:AOB \: and \: COD  } \\  \sf{(i)}&  \sf{ \angle \: OAB =  \angle \: OCD\: (A)}& \sf{AB \parallel \: DC \: and \: alternate \: angles} \\  \sf{(ii)} &\sf{AB = DC(S)}& \sf{Opposite \: sides \: of \: a \: parallelogram} \\  \sf{(iii)} &\sf{ \angle \: OBA=  \angle \: ODC(A)} &\sf{AB \parallel \:DC \: and \: alternate \: angles} \\  \sf{(iv)}& \sf{ \triangle \:AOB\cong \triangle \: COD}& \sf{A.S.A \: axiom}\\ \hline 2.& \sf{AO = OC \: and \: BO = OD}& \sf{Corresponding \: sides \: of \: congruent \: triangle}\\ \hline 3.& \sf{AC \: and \: BD \: bisect \: each \: other \: at \: O}& \sf{From \: statement \: (2)}\\ \\ \hline\end{array}.          Proved ✔

♕ And we're done! Hurrayyy! ;)

# STUDY HARD! So, Tomorrow you can answer people like this , " Dude , I just bought this expensive mobile phone but it is not that expensive for me" [ - Unknown ] :P

☄ Hope I helped! ♡

☃ Let me know if you have any questions! ♪

\underbrace{ \overbrace  {\mathfrak{Carry \: On \: Learning}}} ☂

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

5 0
3 years ago
Can u help me do this
chubhunter [2.5K]
There will be 2 books left over if he fills each box.
7 · 4 = 28
30 - 28 = 2

Good Luck
5 0
3 years ago
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