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LenKa [72]
3 years ago
10

Find 3 consecutive even integers such that the sum of twice the smallest number and 3 times the largest is 42

Mathematics
1 answer:
nadya68 [22]3 years ago
6 0

integers n, n+2 , n+4

2n + 3(n+4) = 42

2n + 3n +12 = 43

5n +12 = 42

5n = 30

n = 6

integers 6, 8 ,10

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<u><em>Longest</em></u>

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8 0
2 years ago
1,Justin rides his bicycle 2.5 kilometers to school. Luke walks 1,950 meters to school.How much farther does Justin ride to scho
kirill [66]
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3 years ago
(b) Ten men can assemble 400 cycles in 8 days. How many cycles 5 men will
Deffense [45]

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8 0
3 years ago
Does anyone know how to solve this? It’s really starting to stress me out.
bekas [8.4K]

Answer:

π − 12

Step-by-step explanation:

lim(x→2) (sin(πx) + 8 − x³) / (x − 2)

If we substitute x = u + 2:

lim(u→0) (sin(π(u + 2)) + 8 − (u + 2)³) / ((u + 2) − 2)

lim(u→0) (sin(πu + 2π) + 8 − (u + 2)³) / u

Distribute the cube:

lim(u→0) (sin(πu + 2π) + 8 − (u³ + 6u² + 12u + 8)) / u

lim(u→0) (sin(πu + 2π) + 8 − u³ − 6u² − 12u − 8) / u

lim(u→0) (sin(πu + 2π) − u³ − 6u² − 12u) / u

Using angle sum formula:

lim(u→0) (sin(πu) cos(2π) + sin(2π) cos(πu) − u³ − 6u² − 12u) / u

lim(u→0) (sin(πu) − u³ − 6u² − 12u) / u

Divide:

lim(u→0) [ (sin(πu) / u) − u² − 6u − 12 ]

lim(u→0) (sin(πu) / u) + lim(u→0) (-u² − 6u − 12)

lim(u→0) (sin(πu) / u) − 12

Multiply and divide by π.

lim(u→0) (π sin(πu) / (πu)) − 12

π lim(u→0) (sin(πu) / (πu)) − 12

Use special identity, lim(x→0) ((sin x) / x ) = 1.

π (1) − 12

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3 0
3 years ago
I’ll give brainliest if u answer within 30 mins
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Answer:

96.7 maybe or 6.75 or 5.67 or even 56.77

Step-by-step explanation:

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