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podryga [215]
2 years ago
5

The difference of twenty-one and eight times a number n is thiry-five

Mathematics
1 answer:
nalin [4]2 years ago
8 0

Answer:

n=7

Step-by-step explanation:

difference means subtraction so:

8n - 21= 35

8n=35+21

8n= 56

n= 56÷8

n=7

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Help me plz :3 :) :x :P
Sonbull [250]
Yes, she is correct
1/2 + 1/2 = 1 or 0.5 + 0.5 = 1 so half of a wall plus half of a wall equals one hole wall.

That's as simple as I can put it mathematical reasoning. I hope this helps
7 0
3 years ago
X-2y+z=2<br> bx+ay=3<br> -2y+z=-4<br> Choose appropriate values for a and b
jonny [76]

Answer:

Attached

Step-by-step explanation:

Just for your reference, this question will get unlimited number of solutions, attached is a few examples I calculated from Excel. Have fun to pick one or create one to be your favorite.

8 0
2 years ago
____ more than 3455 is two hundred seventy -eight thousand five hundred eighty three​
Pavlova-9 [17]
278, 538 is the answer
3 0
3 years ago
Read 2 more answers
If b is the midpoint of ac, ac=cd, ab=3x+4, ac=11x-17, and ce=49, find de
IrinaK [193]

First, we draw our line.

 

|------------------------------------------------------------------------------------|

a                                                                                                            e

 

 

Next, break up this line into segments using the information.

 

|----------------------|----------------------|--------------------|------------------|

a                            b                            c                         d                      e

 

 

The entire line is 29.

 

ab + bc + cd + de = ae

ab + bc + cd + de = 29

 

You also know that

 

bd = bc + cd

 

 

Due to midpoint theorem,

 

ab = bc

cd = de

 

 

Then,

 

2ab + 2cd = 29

 

 

The equations we will use are

 

bd = bc + cd                       eq1

2bc + 2cd = 29                   eq2

 

 

Dividing both sides of the equation in eq2 yields

 

bc + cd = 14.5

 

bd = bc + cd

bd = 14.5

5 0
3 years ago
Read 2 more answers
The recommended dosage for ceclor (cefaclor) is 30mg/kg to be given P.O. for a child weighing 32 lbs. Available: 125mg/5ml. How
andreyandreev [35.5K]

The child's weight in kilogram (kg) is equal to 14.5 kilogram. 2. The safe 24-hour dosage range for this child is 290 mg to 580 mg.

Given that the recommended dosage for ceclor is 30mg/kg to be given P.O. for a child weighing 32 lbs.

In order to determine the weight of the child, we would convert the value of the child's weight in pounds (lbs) to kilogram (kg) as follows:

1 pound = 0.4536 kilogram

32 pound = x kilogram

Cross-multiplying, we have: X = 32 × 0.4536 Child's weight, x = 14.5 kilogram.

Minimum range = 30 × 14.5 = 290 mg to 580 mg

Maximum range = 125 × 14.5 = 580 mg

Therefore, the safe 24-hour dosage range for this child is 290 mg to 580 mg.

Hence based on the calculations, we can infer and logically deduce that yes the dosage ordered safe because the safe single dose range is from 97 mg to 193 mg.

learn more about safe dosage at brainly.com/question/13186642

#SPJ9

Complete Question:

Ceclor (cefaclor) 100 mg p.oh is ordered for a child weighing 32 lbs.

1. What is the child's weight in kilograms (round to tenths)

2. What is safe 24-hour dosage range for this child?

3. Is the dosage ordered safe?

A. Yes

B. No, below the recommended dosage

C. No, it is for above the recommended dosage

D. More information is needed

4 0
1 year ago
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