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vlabodo [156]
3 years ago
9

Please please help please please help me please please help please please

Mathematics
1 answer:
USPshnik [31]3 years ago
5 0

Answer:

Q2) A 0.5//// B 0.11//// C 1.7//// D 1.15//// E 2.13//// F 4.1

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The answer should be A
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Write two equivalent ratios 3to5??
vesna_86 [32]

Answer:

3 : 5

6 : 10

9 : 15

12 : 20

15 : 25

7 0
3 years ago
[HELP ASAP 50 POINTS] 17, 18, 19, 20
Charra [1.4K]

17) f(x) = 16/(13-x).

In order to find domain, we need to set denominator expression equal to 0 and solve for x.

And that would be excluded value of domain.

13-x =0

Adding x on both sides, we get

13-x +x = x.

13=x.

Therefore, domain is All real numbers except 13.



18).f(x) = (x-4)(x+9)/(x^2-1).

In order to find the vertical asymptote, set denominator equal to 0 and solve for x.

x^2 -1 = 0

x^2 -1^2 = 0.

Factoring out

(x-1)(x+1) =0.

x-1=0 and x+1 =0.

x=1 and x=-1.

Therefore, Vertical asymptote would be

x=1 and x=-1

19) f(x) = (7x^2-3x-9)/(2x^2-4x+5)

We have degrees of numberator and denominator are same.

Therefore, Horizontal asymptote is the fraction of leading coefficents.

That is 7/2.


20) f(x)=(x^2+3x-2)/(x-2).

The degree of numerator is 2 and degree of denominator is 1.

2>1.

Degree of numerator >  degree of denominator .

Therefore, there would no any Horizontal asymptote.

6 0
3 years ago
A cup of coffee contains about 100 mg of caffeine. Caffeine is metabolized and leaves the body at a continuous rate of about 12%
Mamont248 [21]

Answer:

a. A = C_{0}(1-x)^t\\x: percentage\ of \ caffeine\ metabolized\\

b. \frac{dA}{dt}= -11.25 \frac{mg}{h}

Step-by-step explanation:

First, we need tot find a general expression for the amount of caffeine remaining in the body after certain time. As the problem states that every hour x percent of caffeine leaves the body, we must substract that percentage from the initial quantity of caffeine, by each hour passing. That expression would be:

A= C_{0}(1-x)^t\\t: time \ in \ hours\\x: percentage \ of \ caffeine\ metabolized\\

Then, to find the amount of caffeine metabolized per hour, we need to differentiate the previous equation. Following the differentiation rules we get:

\frac{dA}{dt} =C_{0}(1-x)^t \ln (1-x)\\\frac{dA}{dt} =100*0.88\ln(0.88)\\\frac{dA}{dt} =-11.25 \frac{mg}{h}

The rate is negative as it represents the amount of caffeine leaving the body at certain time.

3 0
3 years ago
Tisha and her academic team are working to go to state finals. They must have a certain number of points, T, to advance. They ha
creativ13 [48]
I lost my brain cells
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