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natka813 [3]
2 years ago
11

Paula invited some friends to dinner and she plans to make fajitas at the grocery store she spent $2.56 on red bell peppers and

$1.92 on green bell pepper if both types of bell peppers cost $1.60 per pound how many total pounds of bell pepper does she buy
Mathematics
2 answers:
stellarik [79]2 years ago
8 0

Answer:

2.8

Step-by-step explanation:

densk [106]2 years ago
6 0

Answer: 2.8

Step-by-step explanation:

Given Data:

Money spent by Paula for red bell peppers = $2.56

Money spent by Paula on green bell peppers = $1.92

Both pepper cost = $1.60/ pound

Therefore:

Total pounds of bell peppers bought by Paula

= cost of green bell pepper + red red bell pepper / price/pound

= $2.56 + $1.92 / $1.60

= $4.48 / $1.60

= 2.8 pounds

Paula bought 2.8pounds of pepper

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Sophia has an ear infection. The doctor prescribes a course of antibiotics. Sophia is told to take 500 mg doses of the antibioti
Zinaida [17]
Part 1:

Let 

      Q₁ = Amount of the drug in the body after the first dose.
      Q₂ =  250 mg

<span>As we know that after 12 hours about 4% of the drug is still present in the body.
For Q</span>₂<span>,
we get:
             Q</span>₂ = 4% of Q₁ + 250
                   = (0.04 × 250) + 250
                   = 10 + 250
                   = 260 mg

Therefore, after the second dose, 260 mg of the drug is present in the body.

Now, for Q₃ :

<span>We get;
           Q</span>₃ = 4% of Q2 + 250
                = 0.04 × 260 + 250
<span>                = 10.4 <span>+ 250
                = 260.4

</span></span><span><span>For Q₄,
We get;
           Q₄ = 4% of Q₃ + 250 
                 = 0.04 × 260.4 + 250
                 = 10.416 + 250 
                 = 260.416
</span><span>
Part 2:
To find out how large that amount is, we have to find Q₄₀.
Using the similar pattern

for Q₄₀,
We get;
            Q₄₀ = 250 + 250 × (0.04)¹ + 250 × (0.04)² + 250 × (0.04)³⁹

Taking 250 as common;

</span></span>            Q₄₀ =<span> 250 (1 + 0.04 + 0.042 + ⋯ + 0.0439)
<span>                  = 2501 − 0.04401 − 0.04
</span> </span>           Q₄₀ = 260.4167

Hence, T<span>he greatest amount of antibiotics in Susan’s body is 260.4167 mg.

Part 3:
</span>From the previous 2 components<span> of </span>the matter, we all know<span> that </span>the best quantity<span> of the antibiotic in Susan's body is </span>regarding <span>260.4167 mg and </span>it'll<span> occur right </span>once<span> she has taken the last dose. However, </span>we have a tendency to<span> see that already </span>once<span> the fourth dose she had 260.416 mg of the drug in her system, </span>that is simply<span> insignificantly smaller. </span>thus we will<span> say that </span>beginning<span> on the second day of treatment, </span>double every day there'll<span> be </span>regarding<span> 260.416 mg of the antibiotic in her body. Over the course of </span>the subsequent twelve<span> hours {the </span>quantity<span>|the quantity|the number} of the drug </span>can<span> decrease to 4%</span><span> of </span>the most amount<span>, </span>that<span> is 10</span><span>.4166 mg. Then the cycle </span>can<span> repeat.</span>
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Step-by-step explanation:

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Answer:

\dfrac{1}{3}.

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In quadrilateral QRST, Q(-1,0), R(5,0), S(3.5,-6), T(-2.5,-6).

In quadrilateral Q'R'S'T', Q'(-1,2), R'(1,2), S'(0.5,0), T'(-1.5,0).

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D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Using distance formula , we get

QR=\sqrt{(5-(-1))^2+(0-0)^2}=\sqrt{6^2}=6

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½log3 + ½logx

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