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3 years ago

Which ratio does not belong 3:3, 9:12, 12:15, 12:16 Plz Answer for me

Mathematics

2 answers:

ehidna [41]3 years ago

7 0

3:3 doesn't belong because it does not have 12

k0ka [10]3 years ago

3 0

12:16 because others are multiples of 3

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Please help worth lot of points

Nat2105 [25]

It could be y=1 possibly

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3 years ago

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The expression 2(a+b)=−8.1 for certain values of a and b. Find the value of the following expressions for the same values of a a

marusya05 [52]

Answer:

2.025

Step-by-step explanation:

In this problem, you don't necessarily need to find the exact values of a and b. Rather, find the value of (a+b) first, and then you can substitute it into the next equation. First, you can divide both sides of the first equation 2(a+b)= -8.1 by 2, to get (a+b)=-4.05. Now, subsitute the value of (a+b) into the second equation, -0.5(a+b), to get -0.5(-4.05). From here, you can calculate that -0.5 times -4.05 is 2.025. Hope this helped!

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3 years ago

HELP I WILL GIVE BRAINLIEST AND 20 POINTS TO CORRECT ANSWER-Hunter is rolling the dice to see what the probability of rolling di

Neko [114]

Answer:

My assumption of the probability would be..

Step-by-step explanation:

1/2

1/3 or less

1/5 or more

1/8

Because if your looking for probability, Its asking for the chance that it will be able to roll in a certain number so you have a 1/2 of it landing on one or the other side if it's a 2 sided coin or a 1/3 if you have a 3 sided triangle shaped dice.

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3 years ago

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I really really really need help!!!!

Yakvenalex [24]

For $f$ to be continuous at $x=1$ , you need to have the limit from either side as $x\to1$ to be the same.

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(|x-1|+2)=2$
$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(ax^2+bx)=a+b$

If $a=2$ and $b=3$ , then the limit from the right would be $2+3=5\neq2$ , so the answer to part (1) is no, the function would not be continuous under those conditions.

This basically answers part (2). For the function to be continuous, you need to satisfy the relation $a+b=2$ .

Part (c) is done similarly to part (1). This time, you need to limits from either side as $x\to2$ to match. You have

$\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(ax^2+bx)=4a+2b$
$\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(5x-10)=0$

So, $a$ and $b$ have to satisfy the relation $4a+2b=0$ , or $2a+b=0$ .

Part (4) is done by solving the system of equations above for $a$ and $b$ . I'll leave that to you, as well as part (5) since that's just drawing your findings.

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3 years ago

Which of the following would be considered an INDEPENDENT event?

yanalaym [24]

Answer:

A. choosing a sock from a drawer, putting it on, then picking another sock from the drawer

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3 years ago

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