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2 years ago

Please help me solve the problem. Thank you

Mathematics

1 answer:

Afina-wow [57]2 years ago

3 0

9514 1404 393

Answer:

see attached

Step-by-step explanation:

The first attachment shows the rectangles with the area inside and the width on the right. The synthetic division of area by width gives the polynomial coefficients for length. The lengths are marked above the rectangles. The synthetic division is shown the 2nd and 3rd attachments.

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two trains leave the station on time.one train travels north at 60mph .the other train travels south at 50mph.how many miles wil

ozzi

It doesn't matter how far apart just so you know and you answered it 3330 miles apart because you multiply 60 by 50

8 0

3 years ago

) You are finally well!  But now your child is sick!  You know there have been mistakes in orders so you are on the internet t

rjkz [21]

Answer:

The doctor's dosage was not appropriate.

The right dosage is from 56.625 mg to 113.25 mg of Antibiotic to be given every 6 hour for the child with weight 25 lbs.

Step-by-step explanation:

It is given that 20 to 40 mg/kg/day is the recommended dosage.

Now, the doctor's order is 150 mg of antibiotic to be given every 6 hours. And my child weighs 25 lbs.

Now, 1 lb is equivalent to 0.453 kg.

So, my child's weight is (25 × 0.453) = 11.325 kg.

So, the doctor's order is 150 mg of antibiotic to be given every 6 hours for a child of weight 11.325 kg.

Hence, the dosage is [(150 × 4) ÷ 11.325] = 52.98 mg/kg/day.

So, this is not within the limit of 20 to 40 mg/kg/day.

Therefore, the doctor's dosage was not appropriate.

Now, let the appropriate dosage is x mg per every 6 hours for a child with weight 25 lbs i.e. 11.325 kg.

So, $20 \leq \frac{4x}{11.325} \leq 40$

⇒ $20\leq 0.3532x \leq 40$

⇒ $56.625 \leq x \leq 113.25$

So, the right dosage is from 56.625 mg to 113.25 mg of antibiotic to be given every 6 hours for the child with weight 25 lbs. (Answer)

5 0

3 years ago

Find the first four positive values of arccos(0.7). I know how to get the first, but as for the rest, I am lost.

Mandarinka [93]

Now cos⁻¹(0.7) is about 45.6°, that's on the first quadrant.

keep in mind that the inverse cosine function has a range of [0, 180°], so any angles it will spit out, will be on either the I quadrant where cosine is positive or the II quadrant, where cosine is negative.

however, 45.6° has a twin, she's at the IV quadrant, where cosine is also positive, and that'd be 360° - 45.6°, or 314.4°.

now, those are the first two, but we have been only working on the [0, 360°] range.... but we can simply go around the circle many times over up to 720° or 72000000000° if we so wish, so let's go just one more time around the circle to find the other fellows.

360° + 45.6° is a full circle and 45.6° more, that will give us the other angle, also in the first quadrant, but after a full cycle, at 405.6°.

then to find her twin on the IV quadrant, we simply keep on going, and that'd be at 360° + 360° - 45.6°, 674.4°.

and you can keep on going around the circle, but only four are needed this time only.

6 0

3 years ago

Guys help.. this is overdue..

loris [4]

So if his homework takes 25 minutes , and his math takes m minutes , and then as a while it takes h minutes, here’s an example of two equations you can use ;
M+25= h of 25+m=h you can do either one as you please because it doesn’t matter the order as long as they
Both equal h !! Hope this helps , sorry that it’s overdue :( have a great day:))))

4 0

3 years ago

Read 2 more answers

One leg of a right triangle has a length of 3. The other sides have lengths that are consecutive integers. Find these lengths.

Genrish500 [490]

Answer:

Therefore other two sides of the triangle are 4 and 5.

Step-by-step explanation:

Given one leg of a right triangle has a length of 3

Let other sides of the triangle be x and x+1 [ since they are consecutive]

According to the problem,

3²+x²=(x+1)²

⇔9 + x²=x²+2x+1

⇔2x+1=9

⇔2x=9-1

⇔2x=8

⇔x=4

Therefore other two sides of the triangle are 4 and 5.

7 0

3 years ago