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3 years ago

what is the least whole number that has the remainders 10,12, and 15 when divided by 16, 18. and 21, respectively

Mathematics

1 answer:

AnnZ [28]3 years ago

7 0

Answer:

$1002$

Step-by-step explanation:

First find difference between the divisors and remainders.

$16-10=6\\18-12=6\\21-15=6$

Here, the difference between the divisors and remainders is equal.

So, the required number is equal to LCM of $(16,18,21)-6$

$16=2^4\\18=2(3^2)\\21=3(7)$

LCM of $(16,18,21)=2^4(3^2)(7)=1008$

Required Number $=1008-6=1002$

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graph A

no x-intercept

no y-intercept

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*****************************************

Answer: $\frac{150}{31}$

<u>Step by step explanation:</u>

log₆(5x + 6) - log₆(x - 4) = 2

log₆ $(\frac{5x+6}{x-4})$ = 2

$(\frac{5x+6}{x-4})$ = 6²

$(\frac{5x+6}{x-4})$ = 36

5x + 6 = 36(x - 4)

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**********************************************************

Answers:

a. (-1, -5)
b. up
c. x = -1
d. $\frac{-2+\sqrt{15}}{3} and\frac{-2-\sqrt{15}}{3}$
e. -2
f. see attachment
g. domain (-∞, ∞) range [-5, ∞)

<u>Explanation:</u>

f(x) = 3x² + 6x - 2

a=3 b=6 c=-2

x = $\frac{-b}{2a} =\frac{-6}{2(3)} = \frac{-6}{6} = -1$

Axis Of Symmetry: x = -1

f(-1) = 3(-1)² + 6(-1) - 2

= 3 - 6 - 2

= -5

Vertex: (-1, -5)

x = $\frac{-b+/-\sqrt{b^{2}-4ac}}{2a}$

= $\frac{-6+/-\sqrt{6^{2}-4(3)(-2)}}{2(3)}$

= $\frac{-6+/-\sqrt{36+24}}{6}$

= $\frac{-6+/-\sqrt{60}}{6}$

= $\frac{-6+/-2\sqrt{15}}{6}$

= $\frac{-2+/-\sqrt{15}}{3}$

x-intercepts: $\frac{-2+/-\sqrt{15}}{3}$

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