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IgorLugansk [536]
3 years ago
7

Solving Inequalities Independent Practice

Mathematics
1 answer:
Gelneren [198K]3 years ago
3 0

Answer:

X< 9

Step-by-step explanation:

1/3x-6<-3 is correct.

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Evaluate x + y when x = -94 and y = 24<br><br>A: -118<br>B: -70<br>C: 70<br>D: 118​
malfutka [58]
It’s b because 94 is negative so that means you have to subtract to get -70
8 0
3 years ago
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It has been estimated that a student’s final grade in a course decreases by 5 points for every 2 days absent from class. If a st
enyata [817]

Answer:

10 points

Step-by-step explanation:

per 2 days you lose 5 points. To get to 4 days you multiply 2 by 2 and get 4 and 5 by 2 and get 10. So per 4 days he loses 10 points.

5 0
3 years ago
Find the area of the sector in<br> terms of pi.<br> 120°<br> 24<br> Area = [?] π<br> Enter
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Sector area= theta/360 x πr^2

Radius= 24/2= 12

Sector area= 120/360 x π(12)^2

Final answer :

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3 years ago
HELPS PLS 44 and 17 what is y and x
olga_2 [115]
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3 0
3 years ago
If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?
Alisiya [41]
If k is odd, then

\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor

while if k is even, then the sum would be

\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4

The latter case is easier to solve:

\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0

which means k=6.

In the odd case, instead of considering the above equation we can consider the partial sums. If k is odd, then the sum of the even integers between 1 and k would be

S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)

Now consider the partial sum up to the second-to-last term,

S^*=2+4+6+\cdots+(k-5)+(k-3)

Subtracting this from the previous partial sum, we have

S-S^*=k-1

We're given that the sums must add to 2k, which means

S=2k
S^*=2(k-2)

But taking the differences now yields

S-S^*=2k-2(k-2)=4

and there is only one k for which k-1=4; namely, k=5. However, the sum of the even integers between 1 and 5 is 2+4=6, whereas 2k=10\neq6. So there are no solutions to this over the odd integers.
5 0
3 years ago
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