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3 years ago

Part 1: How's the weather?

Mathematics

2 answers:

mariarad [96]3 years ago

8 0

Answer:

Part 1: How's the weather? Investigate the weather forecast in your city.

Record the high and low temperatures that are forecast for the next 7 days in the table below. In the columns titled, Difference," record the change in temperature from the day before. To do this, subtract the previous temperature from the current temperature Be sure to use positive and negative numbers as needed! For example, if yesterday’s temperature was 70 and today’s temperature was 65, the difference is 65 – 70 = -5.

Date

High Temperature

Subtract previous temperature from current temperature

Difference

Low Temperature

Subtract previous temperature from current temperature

Difference

10-24-21

-N/A-

10-25-21

-2

10-26-21

10-27-21

-1

10-28-21

10-29-21

-6

-2

10-30-21

-3

-6

Note: The columns for Day One's subtraction and difference cannot be filled in. As this is the first data point, there is no previous temperature to subtract.

Write a mathematical expression using the information from your table to answer the following questions:

What is the mean change in the forecaster high temperatures? Remember, this can be found by averaging the values in the Difference column for the high temperatures. Show your work and steps. If your answer is not an integer, explain what two integers your answer is between.

What is the mean change in the forecaster low temperatures? Remember, this can be found by averaging the values in the Difference column for the low temperatures. If your answer is not an integer, explain what two integers your answer is between.

Step-by-step explanation:

Rainbow [258]3 years ago

4 0

You do realize nobody’s gonna answer your question right

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kirza4 [7]

A*2 + b* = c*2

(9)*2 + (8)*2

81+64= 145

145=c*2
12.04=c

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3 years ago

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egoroff_w [7]

Answer:

c = -3

Step-by-step explanation:

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$-8=-c-11$

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$-8+11=-c-11+11$

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$3=-c$

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3 years ago

Which of the given functions could this graph represent?

hammer [34]

Answer:

The answer is B. f(x) = (x − 1)(x − 2)(x + 1)(x + 2)

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Given that f(x) = x2 – 14x + 45 and g(x) = 2 – 9, find f(x) – g(x) and

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3 years ago

Find the absolute maximum and minimum values of f(x, y) = x+y+ p 1 − x 2 − y 2 on the quarter disc {(x, y) | x ≥ 0, y ≥ 0, x2 +

Andreas93 [3]

Answer:

absolute max: f(x,y)=1/2+p1 ; at P(1/2,1/2)

absolute min: f(x,y)=p1 ; at U(0,0), V(1,0) and W(0,1)

Step-by-step explanation:

In order to find the absolute max and min, we need to analyse the region inside the quarter disc and the region at the limit of the disc:

Region inside the quarter disc:

There could be Minimums and Maximums, if:

∇f(x,y)=(0,0) (gradient)

we develop:

(1-2x, 1-2y)=(0,0)

x=1/2

y=1/2

Critic point P(1/2,1/2) is inside the quarter disc.

f(P)=1/2+1/2+p1-1/4-1/4=1/2+p1

f(0,0)=p1

We see that:

f(P)>f(0,0), then P(1/2,1/2) is a maximum relative

Region at the limit of the disc:

We use the Method of Lagrange Multipliers, when we need to find a max o min from a f(x,y) subject to a constraint g(x,y); g(x,y)=K (constant). In our case the constraint are the curves of the quarter disc:

g1(x, y)=x^2+y^2=1

g2(x, y)=x=0

g3(x, y)=y=0

We can obtain the critical points (maximums and minimums) subject to the constraint by solving the system of equations:

∇f(x,y)=λ∇g(x,y) ; (gradient)

g(x,y)=K

Analyse in g2:

x=0;

1-2y=0;

y=1/2

Q(0,1/2) critical point

f(Q)=1/4+p1

We do the same reflexion as for P. Q is a maximum relative

Analyse in g3:

y=0;

1-2x=0;

x=1/2

R(1/2,0) critical point

f(R)=1/4+p1

We do the same reflexion as for P. R is a maximum relative

Analyse in g1:

(1-2x, 1-2y)=λ(2x,2y)

x^2+y^2=1

Developing:

x=1/(2λ+2)

y=1/(2λ+2)

x^2+y^2=1

So:

(1/(2λ+2))^2+(1/(2λ+2))^2=1

$\lambda_{1}=\sqrt{1/2}*-1 =-0.29$

$\lambda_{2}=-\sqrt{1/2}*-1 =-1.71$

$\lambda_{2}$ give us (x,y) values negatives, outside the region, so we do not take it in account

For $\lambda_{1}$ : S(x,y)=(0.70, 070)

and

f(S)=0.70+0.70+p1-0.70^2-0.70^2=0.42+p1

We do the same reflexion as for P. S is a maximum relative

Points limits between g1, g2 y g3

we need also to analyse the points limits between g1, g2 y g3, that means U(0,0), V(1,0), W(0,1)

f(U)=p1

f(V)=p1

f(W)=p1

We can see that this 3 points are minimums relatives.

Conclusion:

We compare all the critical points P,Q,R,S,T,U,V,W an their respective values f(x,y). We find that:

absolute max: f(x,y)=1/2+p1 ; at P(1/2,1/2)

absolute min: f(x,y)=p1 ; at U(0,0), V(1,0) and W(0,1)

4 0

3 years ago