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2 years ago

Joe has eaten 2/5 of a pizza.Jane has eaten 1/6 of a pizza how many more pizza has joe eaten than Jane?

1 answer:

3 0

Joe ate 7/30 more of pizza than Jane. You can just subtract 2 uneven denominators. So you would turn them into 30 because that’s the closest highest number that can go with 5 and 6. Then multiply 2 with 6 which is 12 and then multiply 1 with 5 which is 5 now you can subtract it. 12/30-5/30=7/30.

Final Result: 7/30

You might be interested in

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

Translate (in terms of x) then solve the algebraic equation. When the sum of a number and 2 is subtracted from 13 the result is

Diano4ka-milaya [45]

13 - (x+2) = 8
subtract 13 from both sides
-(x+2) = -5
divide by -1 to get rid of negative
(x+2) = 5
subtract 2 from both sides
x=3

7 0

3 years ago

These box plots show the basketball scores for two teams.

Arisa [49]

The most appropriate statement is the interquartile range for the Wolverines, 30 is less than the IQR for the panthers, 40.

<h3>What is the correct statement?</h3>

The box plot is used to show the distribution of data. The box plot can be used to determine the range, interquartile range and median of the data set.

The range is the difference between the two ends of the whiskers.

Range for the Wolverines = 96 - 35 = 61

Range for the Panthers = 107 - 33 = 74

The interquartile range is the difference between the first and third lines on the box

IQR for the Wolverines = 85 - 55 = 30

Range for the Panthers = 90 - 50 = 40

To learn more about box plots, please check: brainly.com/question/1523909

#SPJ1

6 0

2 years ago

Can someone double check that this is correct? If not please correct me :)

bija089 [108]

Yeah you are correct, also I need brainlest please

7 0

2 years ago

You want to find out how sleep deprivation affects motor performance. To study this, you have sleep-deprived subjects (such as p

marysya [2.9K]

Answer:

179.5 - 180.5

Step-by-step explanation:

Time is a continuous variable. The minimum sleep time per night per subject here, is given as 1 minute.

Larger sleep times could be 1.08 minutes, 2.99 minutes, and other continuous/infinite values. Remember there are 60seconds in a minute and in-between seconds, there are milliseconds. So time is a continuous variable.

In this case though, our measurement of time is given in whole number units (integers). Our precision of measurement is 1 unit. We have an observed value of 180 minutes (the first subject's sleep time). The real limits of this value are 179.5 to 180.5

7 0

3 years ago