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3 years ago

What is the solution to this equation?

Mathematics

2 answers:

____ [38]3 years ago

8 0

C is the solution x = 6

labwork [276]3 years ago

6 0

The solution to this equation is C (6)

3x - 3 = 15

3x = 15 + 3
3x = 18
3/3 x = 18/3

x = 6

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Can someone answer this question please answer it correctly if it’s corect I will mark you brainliest

marshall27 [118]

Answer:

5) 27/70

6) 90

Step-by-step explanation:

5) The first step in this problem is to figure out the amount of total spins. To do so, add up all of the numbers in the column "Frequency".

18 + 15 + 27 + 10 = 70.

Now, look at the amount of times the spinner landed on green. This is 27 times. So, the ratio of green spins to total spins is 27:70, or 27 out of 70 spins. Converting this to a fraction, we get the final answer, 27/70.

6) To solve this problem, we have to first do the same steps as the previous problem, but with the color red. There are 70 total spins, and 18 red spins. Therefore, the ratio is 18:70. However, this problem wants the total number of spins to be 350. In other words, 70 needs to become 350. To do this, multiply each side of the ratio by 5. The ratio becomes 90:350. Using this ratio, we can determine that a solid prediction is 90 red spins out of 350 total spins.

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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.

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

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