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MrRissso [65]
2 years ago
7

Do-nothing #1 can count 5 more ceiling dots

Mathematics
1 answer:
klasskru [66]2 years ago
5 0

Answer:

2

Step-by-step explanation:

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What is 2 and 3/4 divided by 1/4
Tanya [424]
2 3/4 divided by 1/4
is
2 3/4 multiplied by 4/1

which is (2+3/4)*4 = 2*4 + 3/4*4 = 8+3 =<em>11</em>
3 0
3 years ago
A 1L IV contains 60meq of kcal. The IV was discontinued after 400ml has infused. How much lvl did the patient receive
Leya [2.2K]

If a 1L IV contains 60meq of kcal and the IV was discontinued after 400ml has infused, then the amount of IV received by the patient is 24meq of kcal.

Calculation for the Amount of IV

It is given that,

1L IV contains 60meq of kcal

⇒ 1000 mL of IV contains 60meq of kcal

⇒ 1 mL of IV will contain 60 / 1000 meq of kcal

As per the question,

The amount of IV infused in the patient = 400 mL

⇒ The amount of IV received by the patient in meq of kcal = 400 × (60 / 1000)

= 4 × 6

= 24 meq of kcal

Hence, the patient receives 24meq of kcal amount of IV.

Learn more about amount here:

brainly.com/question/4567186

#SPJ1

6 0
2 years ago
Which exspression is equivalent to :
dlinn [17]

Answer:

C

Step-by-step explanation:

4 0
4 years ago
Which of the following is an equivalent representation of 5^-4 ?
Afina-wow [57]

Answer:

D

Step-by-step explanation:

5 0
3 years ago
Find the sum of the series Summation from n equals 1 to infinity (StartFraction 3 Over StartRoot n plus 2 EndRoot EndFraction mi
Marizza181 [45]

Looks like the series is supposed to be

\displaystyle\sum_{n=1}^\infty\frac3{\sqrt{n+2}}-\frac3{\sqrt{n+3}}

The series telescopes; consider the kth partial sum of the series,

S_k=\displaystyle\sum_{n=1}^k\frac3{\sqrt{n+2}}-\frac3{\sqrt{n+3}}

S_k=\displaystyle\left(\frac3{\sqrt3}-\frac32\right)+\left(\frac32-\frac3{\sqrt5}\right)+\cdots+\left(\frac3{\sqrt{k+1}}-\frac3{\sqrt{k+2}}\right)+\left(\frac3{\sqrt{k+2}}-\frac3{\sqrt{k+3}}\right)

\implies S_k=\dfrac3{\sqrt3}-\dfrac3{\sqrt{k+3}}

As k\to\infty, the second term converges to 0, leaving us with

\displaystyle\sum_{n=1}^\infty\frac3{\sqrt{n+2}}-\frac3{\sqrt{n+3}}=\frac3{\sqrt3}=\boxed{\sqrt3}}

7 0
3 years ago
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