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34kurt
2 years ago
5

Suraj took a slice of pizza from the freezer and put it in the oven. The pizza's temperature (in degrees Celsius) as a function

of time (in minutes) is graphed.
How long did it take the pizza to reach 0 degrees Celsius?

Mathematics
2 answers:
Molodets [167]2 years ago
6 0
It took about 0.5 minutes to reach level 0
MrRissso [65]2 years ago
3 0

We have been given a graph which describes temperature of pizza ( in degree Celsius) as a function of time (in minutes).

We can see that time is independent variable and temperature is dependent variable as it depends on time.  

Let us find when temperature of pizza will reach 0 degree Celsius. We can find the time when temperature will be 0 degrees by finding when y is zero.  

We can see that initial temperature of pizza is about -6 degree Celsius. As time is increasing temperature of pizza is also getting less negative and it is approaching to zero.        

At t= 0.75 minutes our function h(t) equals zero. Therefore, it will take 0.75 minutes the pizza to reach 0 degree Celsius.


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8.1g of sugar is needed for every cake made. How much sugar is needed for 6 cakes?
Lerok [7]

Answer:

48.6

Step-by-step explanation:

If you use 8.1g of sugar for 1 cake then 6 cakes will be 48.6g of sugar

Just do 8.1*6 and you will get 48.6

8 0
2 years ago
Solve the system of linear equations using multiplication.
Anna71 [15]

Answer:

(8,-1)

Step-by-step explanation:

The given system is:

3x+3y=21

6x+12y=36

Since I prefer to use smaller numbers I'm going to divide both sides of the first equation by 3 and both sides of the equation equation by 6.

This gives me the system:

x+y=7

x+2y=6

We could solve the first equation for x and replace the second x with that.

Let's do that.

x+y=7

Subtract y on both sides:

x=7-y

So we are replacing the second x in the second equation with (7-y) which gives us:

(7-y)+2y=6

7-y+2y=6

7+y=6

y=6-7

y=-1

Now recall the first equation we arranged so that x was the subject. I'm referring to x=7-y.

We can now find x given that y=-1 using the equation x=7-y.

Let's do that.

x=7-y with y=-1:

x=7-(-1)

x=7+1

x=8

So the solution is (8,-1).

We can check this point by plugging it into both equations.

If both equations render true for that point, then we have verify the solution.

Let's try it.

3x+3y=21 with (x,y)=(8,-1):

3(8)+3(-1)=21

24+(-3)=21

21=21 is a true equation so the "solution" looks promising still.

6x+12y=36 with (x,y)=(8,-1):

6(8)+12(-1)=36

48+(-12)=36

36=36 is also true so the solution has been verified since both equations render true for that point.

5 0
2 years ago
Read 2 more answers
Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i
jenyasd209 [6]

Answer:

If the limit that you want to find is \lim_{x\to \infty}\dfrac{P(x)}{Q(x)} then you can use the following proof.

Step-by-step explanation:

Let P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} and Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0} be the given polinomials. Then

\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}

Observe that

\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}

and

\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}

Then

\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}

3 0
2 years ago
Isabella can afford a $410-per-month car payment, and she's interested in either a sedan, which costs $21,600, or a station wago
mylen [45]

Answer:

isabella can afford neither the sedan nor the station wagon

Step-by-step explanation:

you read it and stop being a lazy

5 0
2 years ago
Read 2 more answers
Data from 14 cities were combined for a​ 20-year period, and the total 280 ​city-years included a total of 107 homicides. After
leonid [27]

Answer:

P(0) =  0.6825

P(1) = 0.2607

Step-by-step explanation:

From the given information, the number of homicide is 107 and the total number of homicides per city –year is 280.

Let us denote the number of homicides per city-year as X.

The mean value, X is calculated as:

\begin{array}{c}\\{\rm{Mean}} = \frac{{107}}{{280}}\\\\ = 0.382\\\end{array}  

Mean=   107/280 = 0.382  

The mean number of homicides per city- year \left( {\lambda = \mu } \right)(λ=μ) is 0.382.

a. The probability that zero homicides is obtained is as below:

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ -0.382}}{{\left( {0.382} \right)}^0}}}{{0!}}\\\\ = \frac{{\left( {0.6825 \right)\left( {\rm{1}} \right)}}{1}\\\\ = 0.6825\\\end{array}  

P(X=0) =  e  −0.382  (0.382)⁰​/1  

= (0.6825)(1) /1  

P(X=0) = 0.6825

Thus, the probability that zero homicides P(0) is 0.6825.

b. The probability that one homicides is obtained is as below:

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ -0.382}}{{\left( {0.382} \right)}^0}}}{{0!}}\\\\ = \frac{{\left( {0.6825 \right)\left( {\rm{1}} \right)}}{1}\\\\ = 0.6825\\\end{array}  

P(X=1) =  e  −0.382  (0.382)¹​/1  

= (0.6825)(0.382)/1  

P(X=1) = 0.2607

Thus, the probability that zero homicides P(0) is 0.2607.

​

5 0
2 years ago
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