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3 years ago

1.682 inches rounded to the nearest whole number is 1 inches?

2 answers:

3 0

Answer:

False anything over the halfway point or .5 would be rounded to the number above in this case 1.682 would be rounded to 2.

jok3333 [9.3K]3 years ago

3 0

Answer:

No, it is 2 inches because 6 rounds up.

Hope this helps and have a great day!

And please mark me brainliest if you can :)

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<img src="https://tex.z-dn.net/?f=4b%20-%201%20%3D%20%20-%2013" id="TexFormula1" title="4b - 1 = - 13" alt="4b - 1 = - 13" ali

xz_007 [3.2K]

4b - 1 = -13

4b = -12

b = -3

6 0

3 years ago

What is the area of the trapezoid?

DENIUS [597]

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Answer: 440 ft^2

Explanation:

I hope this helped!

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7 0

2 years ago

Read 2 more answers

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

3 years ago

F(x) =2x-5 and g(x) =x+52 find f(g(x)

irga5000 [103]

If f(x) = 2x - 5 and g(x) = x + 52, then f(g(x)) can be deduced by placing g(x) in the spot of x in the f(x) equation as follows:

f(g(x)) = 2(g(x)) - 5

Since we know g(x) = x + 52, let's plug it in:

f(g(x)) = 2(x + 52) - 5
f(g(x)) = 2x + 104 - 5
f(g(x)) = 2x + 99

4 0

3 years ago

Read 2 more answers

∆ABC rotates around point D to create ∆A'B'C'. However, only is shown. Based on the position of , what are the coordinates of A'

earnstyle [38]

Well if you are talking to me about the following graphic <span>http://contentlaunch.ple.platoweb.com/testimagedb/53/5377183edb46ae5c5177efb0f9dfbbd1 then we have to say that it would be 4,1. Right there you can find A. Hope this has helped you</span>

3 0

3 years ago

Read 2 more answers