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2 years ago

Hello ya'll. I have a quick question to ask ya'll.

Mathematics

1 answer:

stepladder [879]2 years ago

8 0

Answer:

$3n$

Step-by-step explanation:

If the initial number is n, one less than n is n - 1, and one more than n is n + 1. so we have:

$n - 1 + n + n + 1=3n$

At first I thought this answer was too easy, so I tried for n is 6.

$3*6=18\\5+6+7=18$

This is really cool! Hope this helps!

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Adam is subtracting 708 – 10 mentally. He thinks

solong [7]

Adam is correct because when the hundreds and tens digit changes in 708, that means that the 7 and 0 will change into 6 and 9

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2 years ago

Which equation represents a proportional relationship?

iren2701 [21]

The equation that represents a proportional relationship is:

y = 32x

Because:

y/x = 32

So that implies that y is directly proportional to x.

Note that other equations represent a proportional relationship in some form, only that the relationship y = 32x is the only direct proportional relationship.

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2 years ago

Pat is a 23 year old man. He earns a taxable income of $70,000.00, and the effective

Jobisdone [24]

Answer:

$12,594.71

Step-by-step explanation:

i did the test

3 0

2 years ago

What line is parallel y=5x-3

Korvikt [17]

Im pretty sure its y= 5x-2

4 0

2 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