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xxTIMURxx [149]
3 years ago
13

I need help please!!

Mathematics
1 answer:
Semmy [17]3 years ago
5 0

Answer:

Step-by-step explanation:

5.   <2=<10=102

<3=<9=78

<4=<10=102

<5=<7=38

<6=180-38=142

<8=142

<9=180-102=78

<11=<9=78

<12=<10=102

<13=<7=38

<14=<16=<6=<8=180-38=142

<15=<7=38

<16=142

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Global mean temperature was changing the most rapidly in 2005. How quickly was the temperature changing in 2005
AysviL [449]

The temperature was changing at a rate of 2.925 degrees Celsius in 2005

<h3>How to determine the rate in 2005</h3>

From the graph (see attachment), we have the following ordered pair

(x,y) = (5,14.625)

The above means that, the temperature in 2005 is 14.625 degrees Celsius

So, the rate in 2005 is:

Rate = \frac{14.625}{5}

Rate = 2.925

Hence, the temperature was changing at a rate of 2.925 degrees Celsius in 2005

Read more about average rates at:

brainly.com/question/8728504

4 0
2 years ago
Integrate this two questions ​
tankabanditka [31]

Simplify the integrands by polynomial division.

\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)

\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)

Now computing the integrals is trivial.

5.

\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}

where we use the power rule,

\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)

and a substitution to integrate the last term,

\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)

8.

\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}

using the same approach as above.

5 0
2 years ago
Please help me with this!
morpeh [17]

Answer:

Step-by-step explan

6 0
2 years ago
Which statement is true about the expressions below?
Zielflug [23.3K]
Is there a attachment because I don’t see anything
7 0
3 years ago
Can someone help me please
iris [78.8K]

Answer:

"I can find the maximum or minimum by looking at the factored expression of a quadratic function by reading off its roots and taking the arithmetic average of them to obtain the x-coordinate of the quadratic function, and then substituting that value into the function."

Step-by-step explanation:

Because of the symmetry of quadratics (which is the case here because we have two factors of degree 1, so we are dealing with a <em>polynomial</em> of degree 2, which is a fancy way of saying that something is a quadratic), the x-coordinate of the extremum (a fancy way of saying maximum or minimum) is the (arithmetic) average of the two roots.

In the factored form of a quadratic function, we can immediately read the roots: 3 and 7. Another way to see that is to solve (x-3)(x-7)=0, which gives x-3=0 \vee x-7=0 (the 'V' stands for 'or'). We can take the average of the two roots to get the x-coordinate of the minimum point of the graph (which, in this case, is x=\frac{3+7}{2} = \frac{10}{2} = 5).

Having the x-coordinate of the extremum, we can substitute this value into the function to obtain the maximum or minimum point of the graph, because that will give the y-coordinate of the extremum.

6 0
3 years ago
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