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Kipish [7]
3 years ago
7

Solve0.9(7x + 14) = 1.5 - (x + 2)

Mathematics
1 answer:
Triss [41]3 years ago
7 0

\\ \sf\longmapsto 0.9(7x+14)=1.5-(x+2)

\\ \sf\longmapsto 6.3x+12.6=1.5-x-2

\\ \sf\longmapsto 63x+12.6=-x-0.5

\\ \sf\longmapsto 63x+x=-0.5-12.6

\\ \sf\longmapsto 64x=-13.1

\\ \sf\longmapsto x=\dfrac{-13.1}{64}

\\ \sf\longmapsto x=0.2

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in triangle ABC shown below, L is the midpoint of BC, M is the midpoint of AB, and N is the midpoint of AC. Find the perimeter o
alexdok [17]

35

since m and n are the midpoints of ab and ac resistively

then ,bc=2mn=16

since m and l are the midpoints of ba and bc respectively

then, ac =2mn =10

then,nc =1/2 × 10 =5

similary, mb=6

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6 0
3 years ago
23 and 24 I need help please
Gnoma [55]

Answer:

They should purchase the snack with the purple box. This is the answer for 23, i don't know what 24 is sorry.

Step-by-step explanation:

24÷3=8

8×5=40

so the first snacks would cost them $40.

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5 0
3 years ago
Is 7/8 or 11/10 closer to 1? How did you decide anybody help please?
erik [133]

Let

A-----> \frac{7}{8}

B-----> \frac{11}{10}

C-----> 1

we know that

Point A-------> Multiply numerator and denominator by 10

\frac{7*10}{8*10} =\frac{70}{80}

Point B-------> Multiply numerator and denominator by 8

\frac{11*8}{10*8} =\frac{88}{80}

Point C-------> Multiply numerator and denominator by 80

\frac{1*80}{1*80} =\frac{80}{80}

Find the distance Point A to Point C

\frac{80}{80} -\frac{70}{80} =\frac{10}{80}

Find the distance Point B to Point C

\frac{88}{80} -\frac{80}{80} =\frac{8}{80}

therefore

\frac{8}{80} < \frac{10}{80}

Point B is closer to Point C

\frac{11}{10} is closer to 1

the answer is

\frac{11}{10} is closer to 1

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