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aliya0001 [1]
3 years ago
9

Please help on two math questions. I don’t understand how to do them.

Mathematics
1 answer:
Sidana [21]3 years ago
3 0

When you represent intervals on the number line, you're including full dots, excluding empty dots, and you're considering numbers highlighted by the line.

In the first case, you've highlighted everything before -2 (full dot, thus included), and everything after 1 (empty dot, excluded). So, the set would be

x\leq -2 \lor\ x>1

or, in interval notation,

(-\infty,-2]\cup (1,\infty)

In the second case, you are looking for all numbers between -3 and 5. This interval is symmetric with respect to 1: you're considering all numbers that are at most 4 units away from 1, both to the left and to the right.

This means that the difference between your numbers at 1 must be at most 4, which is modelled by

|x-1|\leq 4

where the absolute values guarantees that you'll pick numbers to the left and to the right of 1.

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Step-by-step explanation:

Given that a  bank representative studies compound interest, so she can better serve  customers. She analyzes what happens when $2,000 earns interest several  different ways at a rate of 2% for 3 years.

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e) the interest if it is compounded monthly.

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3 years ago
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Pachacha [2.7K]

Answer:

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Step-by-step explanation:

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2 years ago
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Landon and Vanessa are at an arcade. Landon scored 48,827 points on a video game. Vanessa scored 62,919 on the same game. What w
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3 years ago
3x+-2y=-6<br> 5x-2y=7<br> how do i solve this Elimination pls
tatuchka [14]

Answer:

x=\frac{13}{2}, y=\frac{51}{4}

Step-by-step explanation:

We\ are\ given,\\3x-2y=-6\\5x-2y=7\\Considering\ this\ system\ of\ equations,\\Let\ 3x-2y=-6\ be\ E_1\\Let\ 5x-2y=7\ be\ E_2,\\Hence,\\Multiplying\ E2\ with\ -1, we\ get:\\3x-2y=-6\\-1(5x-2y)=7*-1\\Hence,\\3x-2y=-6\\-5x+2y=-7\\Now,\\Lets\ add\ E_1\ and\ E_2\ together\\Hence,\\(3x-2y)+(-5x+2y)=(-6)+(-7)\\Hence,\\3x-2y-5x+2y=-13\\Arranging\ And\ Adding\ Like\ terms,\ we\ have:\\3x-5x-2y+2y=-13\\Hence,\\-2x+0y=-13\\-2x=-13\\x=\frac{-13}{-2}=\frac{13}{2}

Now,\ we\ can\ substitute\ x=\frac{13}{2}\ in\ E_1\ or\ E_2,\\But,\\Here,\ we'll\ substitute\ it\ in\ E1,\\Hence,\\Substituting\ x=\frac{13}{2}\ in\ E1,\ we\ have:\\3*\frac{13}{2}-2y=-6\\\frac{39}{2}-2y=-6\\-2y=-6- \frac{39}{2}\\-2y=\frac{-12-39}{2}\\-2y=\frac{-51}{2}\\y=\frac{-51}{2*-2}=\frac{51}{4}\\Hence,\\x=\frac{13}{2}, y=\frac{51}{4}

4 0
3 years ago
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