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deff fn [24]
2 years ago
8

To control an infection, a doctor recommends that a patient who weighs 92 pounds be given 320 milligrams of antibiotic. If the a

ntibiotic is given proportionally according to the patient's weight, how much antibiotic should be given to a patient who weighs 138 pounds? 400 milligrams 480 milligrams 550 milligrams 600 milligrams
Mathematics
2 answers:
aleksandrvk [35]2 years ago
7 0

Answer: 480 milligrams

Step-by-step explanation:

We can solve this by using direct proportion whereby y = kx

where,

x = patient's weight = 92 pounds

k = constant

y = milligrams of antibiotics = 320

Using y = kx

320 = 92k

k = 320/92

k = 3.4782609

The antibiotic that should be given to a patient who weighs 138 pounds will be:

y = kx

y = 3.4782609 × 138

y = 480 milligrams

The antibiotic that should be given to a patient who weighs 138 pounds is 480 milligrams.

loris [4]2 years ago
4 0

Anwser: <u>B - 480 milligrams</u>

Why:

Because I just took the test on Edgen.uity

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Could someone please help on this question? I’m giving 20 points and brainliest but I need it quickly so please help!
AnnZ [28]
<h3>Answer:  HF = 5</h3>

=============================================

Explanation:

Refer to the diagram below. I'm adding in points A and B, which are marked in blue. I'm also adding in variables x, y, and z.

The goal is to find HF, so let's make that equal to x. This is also equal to HB because they are radii of the same circle.

Let y be the length of segments JA and JB.

Let z be the length of segments CF and CA.

So in short we have

  • x = HF = HB
  • y = JA = JB
  • z = CF = CA

We know that

  • CH = 13
  • HJ = 19
  • CJ = 22

This must mean

  • CH = CF+HF = z+x = 13
  • HJ = HB+JB = x+y = 19
  • CJ = CA+JA = z+y = 22

Or in short,

  • z+x = 13
  • x+y = 19
  • z+y = 22

Let's solve the first equation for z to get z = -x+13. I subtracted x from both sides.

Now plug this into the third equation to get

z+y = 22

(z) + y = 22

(-x+13) + y = 22 ... replace z with -x+13

-x+13+y = 22

-x+y+13 = 22

-x+y = 22-13 .... subtract 13 from both sides

-x+y = 9

------------------------------------

So we have this reduced system of equations of two variables and two equations

  • x+y = 19 ... found earlier
  • -x+y = 9 .... what we just found above

Add the equations straight down. This is the elimination property.

Doing so leads to the x terms canceling out since x + (-x) = 0x = 0

The y terms add to y+y = 2y

The terms on the right hand side add to 19+9 = 28

We are left with 2y = 28 which solves to y = 28/2 = 14.

If y = 14, then,

x+y = 19

x+14 = 19

x = 19-14

x  = 5

In which we can then say

z = -x+13

z = -5+13

z = 8

Though we don't need to find z (unless you're curious about it).

Since x = 5, and we set HF equal to x, this means that HF = 5.

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3 years ago
Value 23.7 is 1/10 of
inna [77]
Is 19:17 because u have to value the number of the first like this 78:807y :997
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If the probability of getting an odd number is 5/10 on a game spinner, the numerator of the practice is:
Tom [10]
The numerator is 5 (and the denominator is 10).
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BaLLatris [955]

Here , we are provided with a table which shows 5 consecutive terms of an arithmetic sequence . But before solving further , let's recall that ;

The n'th term of a Arithmetic Sequence let's say it be {\sf T_n} is given by ;

  • {\boxed{\bf T_{n}=T_{1}+(n-1)d}}

Where , <u>d</u> is the common difference

Now , here we are given with ;

{\quad \qquad \sf \blacktriangleright \blacktriangleright \blacktriangleright T_{1}=8 \: and \: T_{5}=-4}

We have to find the 2nd , 3rd and 4th term respectively ,

Now , by using the above formula , 5th term can be written as ;

{: \implies \quad \sf T_{1}+(5-1)d=T_{5}}

Putting the values and transposing 1st term to RHS , we have ;

{: \implies \quad \sf 4d = -4-8}

{: \implies \quad \sf d=-\dfrac{12}{4}}

{: \implies \quad \sf d=-3}

Now , as we got the common difference , so we can find out the missing terms now ;

{: \implies \quad \sf T_{2}= T_{1}+(2-1)d}

{: \implies \quad \sf T_{2}= 8 +d}

{: \implies \quad \sf T_{2}= 8-3}

{: \implies \quad \bf \therefore \:  T_{2}= 5}

Now

{: \implies \quad \sf T_{3}= T_{1}+(3-1)d}

{: \implies \quad \sf T_{3}= 8 +2d}

{: \implies \quad \sf T_{3}= 8-6}

{: \implies \quad \bf \therefore \:  T_{3}= 2}

Also ,

{: \implies \quad \sf T_{4}= T_{1}+(4-1)d}

{: \implies \quad \sf T_{4}= 8 +3d}

{: \implies \quad \sf T_{4}= 8-9}

{: \implies \quad \bf \therefore \:  T_{4}= -1}

Now , The given table can be written as ;

{\begin{array}{|c|c|c|c|c|c|}\cline{1-6} \bf n & \sf 1 & 2 & 3 & 4 & 5 \\ \cline{1-6} \bf T_{n} & \sf 8 & 5 & 2 & -1 & -4 \end{array}}

Note :- Kindly view the answer from web , if you're not able to see the full answer from here ;

brainly.com/question/26750175

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2 years ago
Can someone please help me asap !!!!!!.???
Tatiana [17]
A, C, D, F? Not sure but took a good guess
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3 years ago
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