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4 years ago

5

Each tablet is 250 milligrams. Your doctor wants you to take 3 grams (g) per day. How many tablets do you need to take per day?

What are the steps to solve this?

1 answer:

liraira [26]4 years ago

4 0

Each tablet is 250 milligrams. Your doctor wants you to take 3 grams (g) per day

We know ,

1 gram = 1000 milligrams

250 milligrams makes 1 tablet

so 1000 milligrams makes how many tablets

$\frac{1000}{250}$ = 4 tablets

1000 milligrams makes 4 tablets. That is 1 gram = 4 tablets

1 gram = 4 tablets

3 grams = 3 * 4 = 12 tablets

So you need to take 12 tablets per day.

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loris [4]

Answer:

30 i think

Step-by-step explanation:

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a rabit can run 35 miles per hour.a fox can run 21 miles in half an hour. which animal is faster, an by how much

Ilia_Sergeevich [38]

Answer:
The fox is faster, by 7 miles per hour

Step-by-step explanation:

The fox can run 21 miles in half an hour -> It can run 42 miles in an hour

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-Solve the proportion (round to the nearest tenth).<br><br> 1910=x12

zysi [14]

The value of x in the proportion is 15.8.

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read more: brainly.com/question/14026746?referrer=searchResults

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3 years ago

When I visit my brother it is a 532 mile round trip. If I visit my brother once a month how many miles would I travel in a year

almond37 [142]

Answe12800 miles

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3 years ago

Let R be the region bounded by

loris [4]

a. The area of $R$ is given by the integral

$\displaystyle \int_1^2 (x + 6) - 7\sin\left(\dfrac{\pi x}2\right) \, dx + \int_2^{22/7} (x+6) - 7(x-2)^2 \, dx \approx 9.36$

b. Use the shell method. Revolving $R$ about the $x$ -axis generates shells with height $h=x+6-7\sin\left(\frac{\pi x}2\right)$ when $1\le x\le 2$ , and $h=x+6-7(x-2)^2$ when $2\le x\le\frac{22}7$ . With radius $r=x$ , each shell of thickness $\Delta x$ contributes a volume of $2\pi r h \Delta x$ , so that as the number of shells gets larger and their thickness gets smaller, the total sum of their volumes converges to the definite integral

$\displaystyle 2\pi \int_1^2 x \left((x + 6) - 7\sin\left(\dfrac{\pi x}2\right)\right) \, dx + 2\pi \int_2^{22/7} x\left((x+6) - 7(x-2)^2\right) \, dx \approx 129.56$

c. Use the washer method. Revolving $R$ about the $y$ -axis generates washers with outer radius $r_{\rm out} = x+6$ , and inner radius $r_{\rm in}=7\sin\left(\frac{\pi x}2\right)$ if $1\le x\le2$ or $r_{\rm in} = 7(x-2)^2$ if $2\le x\le\frac{22}7$ . With thickness $\Delta x$ , each washer has volume $\pi (r_{\rm out}^2 - r_{\rm in}^2) \Delta x$ . As more and thinner washers get involved, the total volume converges to

$\displaystyle \pi \int_1^2 (x+6)^2 - \left(7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \pi \int_2^{22/7} (x+6)^2 - \left(7(x-2)^2\right)^2 \, dx \approx 304.16$ <em />

d. The side length of each square cross section is $s=x+6 - 7\sin\left(\frac{\pi x}2\right)$ when $1\le x\le2$ , and $s=x+6-7(x-2)^2$ when $2\le x\le\frac{22}7$ . With thickness $\Delta x$ , each cross section contributes a volume of $s^2 \Delta x$ . More and thinner sections lead to a total volume of

$\displaystyle \int_1^2 \left(x+6-7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \int_2^{22/7} \left(x+6-7(x-2)^2\right) ^2\, dx \approx 56.70$

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