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Brilliant_brown [7]

3 years ago

15

Anyone know this. Pls help now.

1 answer:

Annette [7]3 years ago

8 0

Scale factor basically means multiplying.

New coordinates: -12, 0

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What is the answer to this math question!?

Nostrana [21]

Answer:

C) 37

Step-by-step explanation:

1/2(PQ)=RQ

2.5x+17=6x-11

17=3.5x-11

28=3.5x

x=8

RQ=6(8)-11

RQ=48-11

RQ=37

5 0

2 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=55%20%5Cfrac%7B6%7D%7B9%3F%7D%20%20%5Ctimes%2066%20%5Cfrac%7B5%7D%7B6%7D%20" id="TexFormula1"

JulsSmile [24]

Answer:

Step-by-step explanation:

If you can present a problem in Latex, you can do anything. I don't know what the question mark is for. I'm just ignoring it.

55 2/3 * 66 5/6

One of the ways to get the answer is to use decimals

55.666666667 * 66.833333333 = 3720.38889

Another way to do this problem is to break up one of the numbers

55 2/3 (66 + 5/6) You can do this if you know how to use the distributive property.

55 2/3 * 66 + 55 2/3 * 5/6

( (165 + 2) / 3) * 66 + (165 + 2)/3 * 5/6

167/3 * 66 + 167 / 3 * 5/6

167 * 22 + (167 * 5 / (3 * 6)

3674 + 835 / 18

3674 + 46 7/18

3720 7/18

If none of these seem right and you have choices, please list them.

6 0

3 years ago

The table shows the number of hours spent practicing singing each week in three samples of 10 randomly selected chorus members.

astraxan [27]

Answer:

B they are close together

Step-by-step explanation:

6 0

2 years ago

37/8 as a mixed number

Alexxandr [17]

4 5/8
you see how many time 8 goes into 37 which is 4 and then you have 5 left over

8 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

2 years ago