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2 years ago

How do you graph 16x-4y=2? Please give me bigger numbers then (1/8,0) , (0,-1/2)

Mathematics

1 answer:

OverLord2011 [107]2 years ago

3 0

Answer:

x= ( 0, 1/8)

y= ( -1/2, 0)

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SOMEONE PLEASE PLEASE HELP ME

slava [35]

Answer:

taking cos

co62=b/h

cos 62=x/12

0.46947x12=x

so, x = 5.633

rounding off

x= 6

6 0

2 years ago

Read 2 more answers

8. A concert charges $10 for each adult ticket and $5 for each child ticket. Sunday, the park collected a total of $1,750

Sphinxa [80]

Do you need both of the equations or just the entire solution?

6 0

2 years ago

Ray is six years younger than Tom. John is three times as old as Ray. Let Tom be x years old. The three of them

gtnhenbr [62]

Answer:

Ray = 19 yrs old
Tom = 25 yrs old
John = 57 yrs old

Explanation:

Let x be the years of Tom

R = x - 6
J = 3R = 3(x - 6)
T = x

R + J + T = 101
x - 6 + 3(x - 6) + x = 101
2x - 6 + 3x - 18 = 101
5x - 24 = 101
5x = 125
x = 125/5 = 25

R = 25 - 6 = 19
J = 3(25 - 6) = 57
T = 25

6 0

2 years ago

What is the slope of a line through (-3, 4) and<br> (5, 6)?

Likurg_2 [28]

Answer:

slope= x2- x1/y2 -y1

= 5-(-3)/6-4

8/2= 4

7 0

2 years ago

Find the maximum and minimum values attained by f(x, y, z) = 5xyz on the unit ball x2 + y2 + z2 ≤ 1.

Allushta [10]

Check for critical points within the unit ball by solving for when the first-order partial derivatives vanish:
$f_x=5yz=0\implies y=0\text{ or }z=0$
$f_y=5xz=0\implies x=0\text{ or }z=0$
$f_z=5xy=0\implies x=0\text{ or }y=0$

Taken together, we find that (0, 0, 0) appears to be the only critical point on $f$ within the ball. At this point, we have $f(0,0,0)=0$ .

Now let's use the method of Lagrange multipliers to look for critical points on the boundary. We have the Lagrangian

$L(x,y,z,\lambda)=5xyz+\lambda(x^2+y^2+z^2-1)$

with partial derivatives (set to 0)

$L_x=5yz+2\lambda x=0$
$L_y=5xz+2\lambda y=0$
$L_z=5xy+2\lambda z=0$
$L_\lambda=x^2+y^2+z^2-1=0$

We then observe that

$xL_x+yL_y+zL_z=0\implies15xyz+2\lambda=0\implies\lambda=-\dfrac{15xyz}2$

So, ignoring the critical point we've already found at (0, 0, 0),

$5yz+2\left(-\dfrac{15xyz}2\right)x=0\implies5yz(1-3x^2)=0\implies x=\pm\dfrac1{\sqrt3}$
$5xz+2\left(-\dfrac{15xyz}2\right)y=0\implies5xz(1-3y^2)=0\implies y=\pm\dfrac1{\sqrt3}$
$5xy+2\left(-\dfrac{15xyz}2\right)z=0\implies5xy(1-3z^2)=0\implies z=\pm\dfrac1{\sqrt3}$

So ultimately, we have 9 critical points - 1 at the origin (0, 0, 0), and 8 at the various combinations of $\left(\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3}\right)$ , at which points we get a value of either of $\pm\dfrac5{\sqrt3}$ , with the maximum being the positive value and the minimum being the negative one.

5 0

2 years ago