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katovenus [111]

3 years ago

9

Which is greater 4/5 or 60% i need help

2 answers:

Kisachek [45]3 years ago

5 0

Answer:

4/5 is greater

Step-by-step explanation:

60% = 60/100

4/5 = 80/100

80>60

BartSMP [9]3 years ago

4 0

4/5= 80%, so 4/5.
It’s pretty easy once you get the hang of it

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Chang knows one side of a triangle is 13 cm. Which set of two sides is possible for the lengths of the other two sides

marta [7]

Answer:

8cm and 9cm

Step-by-step explanation:

because for the triangle to work you need the other two other sides when added together needs to be greater then 13

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2 years ago

The ratio of the side length of square a to the side length of square b is 3:5. the perimeter of square b is 60 feet. what is th

marishachu [46]

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2 years ago

Which r-value suggests a weak positive correlation? A. r = 0.17454 B. r = −0.17454 C. r = 0.98264 D. r = −0.98264

Vaselesa [24]

Answer:

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Step-by-step explanation:

Got it right on my text

5 0

2 years ago

-5+4y=-21 simplify your answer as much as possible <br>y=

nordsb [41]

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simplified

3 0

3 years ago

Read 2 more answers

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