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Anna11 [10]
3 years ago
12

How to start solving it

Mathematics
1 answer:
katrin [286]3 years ago
3 0
It is equal to - 11/20 is the same as 55%...
Hope this helped!
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Sarah's Shipping and Ryan's Mail Services both ship packages. Sarah's trucks
algol [13]

Answer:I think 11

Step-by-step explanation: 11 because one can do no more than 11 packages so the most potential one most be 11 since it can't do 18 but the 18 can do 11 and the max of the other one is 11

8 0
2 years ago
A carpenter has less than 120 min. to spend painting furniture each day. Today, he has spent 30.5 min. painting a desk. Now he w
FinnZ [79.3K]

Answer:

Hence, a carpenter can paint a maximum of 7 chairs.

Hence, option A is correct.

Step-by-step explanation:

A carpenter has less than 120 min. to spend painting furniture each day.

Today, he has spent 30.5 min. painting a desk.

Now he will paint x chairs, each of which takes 12.5 min.

i.e. he will take a total of 12.5x minutes in painting 'x' chairs.

Also total time spent on painting furniture each day will be equal to time spent on painting a desk and all of the chairs.

Hence,

30.5+12.5x<120 (since the total time he spends on painting is less than 120 mins.)

⇒ 12.5 x<120-30.5

⇒ 12.5 x<89.5

⇒ x<7.16  ( on dividing both side of the inequality with 12.5)

Hence, a carpenter can paint a maximum of 7 chairs.


6 0
3 years ago
Read 2 more answers
How many units are -7 and its opposite from zero?
kenny6666 [7]

Answer:

-7 is -7 units away from zero. Its opposite is 7 from zero.

Step-by-step explanation: Count from 0 how many digits away it is

3 0
1 year 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
Identify the pair of angles shown in the figure.
Katen [24]

Answer:

d

Step-by-step explanation:

i think

7 0
2 years ago
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