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Sladkaya [172]
2 years ago
10

Last year Jeremy weighed 70 bounds this year he weights 84 pounds by what percent did Jeremy weights increase from last year to

this year
Mathematics
1 answer:
stiks02 [169]2 years ago
6 0
14%I think hopefully
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a group of 25 sixth graders, 40% of them said they had never visited another state. How many students said they had never visite
diamong [38]
Correct Answer: 10 students have never visited another state
40/100 × 25 =
40 is out of 100 because percentages always is represented by 100 (cent)

cross out the zero's equally top and bottom 
so it would become: 4/100 × 25 = 4 ×25 = 100
 then add the denominator
100/10 is our final resolution 

cross out the zeros evenly and answer will give you 10.
6 0
3 years ago
Read 2 more answers
What value of y makes the equation true. y+2.9=11
yawa3891 [41]

Answer:

y = 8.1

Step-by-step explanation:

y+2.9=11

Subtract 2.9 from each side

y+2.9-2.9 = 11-2.9

y =8.1

8 0
3 years ago
I need help ASAP PLEASEE
Alex787 [66]

Answer:

9. 21

10. 70

11. 64

12. 31

13. 85

14. 31

double check the answer, i didnt use a calculator so might be wrong

4 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
Please help!! Math II
Vaselesa [24]
I say that x equals 5
7 0
3 years ago
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