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3 years ago

How do I find the simplest form,for this problem 4-2(3+6)+2-2=

1 answer:

5 0

Try using PEMDAS,
Parentheses
Exponents
Multiplication
Division
Addition
Subtraction

So just do the problem in that order.

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3 years ago

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Write the fraction as a sum of fraction three different ways 7/10

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Step-by-step explanation:j7/10. 7 divided by 10. And .7

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terrence buys a new car for $20,000. the value of the car depreciates by 15% each year. if f(x) represents the value of the car

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3 years ago

PLS HELP ME FIND THE MAD, MADIAN, RANGE AND IQR OF THIS DATA SET.

sweet-ann [11.9K]

Answer:

The median is 6. The Mean Absolute Deviation is 1.88. The range is 10. I believe the IQR is 5.

Step-by-step explanation:

the median is just the number that falls in the center of a data set. The Mean Absolute Deviation is the distance of the data points from the mean. So, find the sum of the differences of each data point from the mean and divide it by the number of data points. Range is the difference between the smallest and largest values in a data set. The interquartile range is the middle of the data. it is a measure of how widely the middle half of the data is spread around the median.

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2 years ago

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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.

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3 years ago