Ask question

Ask question

All categories

3 years ago

11

I need help I dont really get fractions lol. I will mark brainliest!

2 answers:

nevsk [136]3 years ago

8 0

ANSWER:
7/8 and 6/8

7/8 remains the same

3/4 multiply the top and bottom by 2 so
3*2=6
4*2=8
so its so 6/8

Reptile [31]3 years ago

3 0

Answer is c 7/8 and 6/i

You might be interested in

A hotel room cost $79 per night. Meg's family is getting 2 rooms for 5 nights for their trip.

SCORPION-xisa [38]

$790 because it would be 79 times 10

6 0

2 years ago

Please help 2 qeastions

Neko [114]

B and D ..............

3 0

3 years ago

Read 2 more answers

Find the distance between-10,-13 and 20,-13

Ray Of Light [21]

Answer:

the distance between these 2 would be 30 units

Step-by-step explanation:

Lets ignore the -13 for now because it doesn't have anything to do with the question

Now simply just add 10+ 20 (i took out the negative because it doesn;t make sense)

10+20

=30

plz brainliest :'D

4 0

3 years ago

10.204 round to the nearest hundredth

Nonamiya [84]

Answer:

10

Step-by-step explanation:

5 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

5 0

3 years ago