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scoray [572]
2 years ago
6

Choose the inequality that describes the following problem.

Mathematics
1 answer:
Yuki888 [10]2 years ago
3 0

Answer:

x\geq 125

Step-by-step explanation:

  • So first we have to find out the total number of points possible
  • So he took 5 tests each with a max score of 100 points, so that's 500 points total
  • The final test will be 150 points, so his whole grade is out of 500+150= 650 points
  • He has already scored 460 of those 650 points
  • Now how much must he get in the least, on the last test in order to make 90%?
  • Now we need to find what 90% of 650 points is
  • so \frac{90}{100} *650=\frac{90*650}{100} =585points
  • Then we ask ourselves, how much must we add to 460, to get 585 points? 585-460= 125 points
  • So Mark needs to get either 125 points or more on the next test to make 90% or more
  • Now let's say the letter x means whatever grade number Mark gets on his next test
  • This number must not be less than 125, so it must be either equal to or greater than 125 points
  • So x\geq 125
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Let Z={a,c,{a,b}}. What is |Z|?
Lina20 [59]

Z=\{a,c,\{a,b\}\}

\boxed{|Z|=3} (treat \{a,b\} as one element of Z)

The power set of Z is

\boxed{2^Z=\bigg\{\{\},\{a\},\{c\},\big\{\{a,b\}\big\},\{a,c\},\big\{a,\{a,b\}\big\},\big\{c,\{a,b\}\big\},\big\{a,c,\{a,b\}\big\}\bigg\}}

1. \{a,c\}\subseteq Z is true because both a\in Z and c\in Z.

2. a\in Z is true.

3. \{c\}\subseteq Z is true (same reason as part 1).

4. \{c\}\in Z is false because Z does not contain the set \{c\}, rather just the element c itself.

5. b\in Z is false because the element b on its own simply is not in Z. That b\in\{a,b\} does not mean b\in Z, but that b belongs to a subset of Z.

6. \{a,b\}\in Z is true.

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3 years ago
I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS AND EXPLAIN WHY THAT IS THE ANSWER
ivanzaharov [21]

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Step-by-step explanation:

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Could someone explain how to do this, I have no idea how...
alexira [117]

Answer:

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Step-by-step explanation:

Pair 'like' terms with 'like' terms, ie numbers go with numbers, and 'a's go with 'a's.

Lets deal with the top of the fraction first:

4ax3a³

Rearrange it so you have numbers beside numbers and 'a's beside 'a's:

(4x3)x(axa³)

12x(a⁴) <em>(because nᵃxnᵇ=nᵃ⁺ᵇ)</em>

12a⁴

Now, instead of (4ax3a³)/6a², we have 12a⁴/6a²

First divide the numbers: 12/6 =2

Now divide the 'a' parts: a⁴/a²=a² <em>(because nᵃ/nᵇ=nᵃ⁻ᵇ)</em>

Now we have 2a²

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2 years ago
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