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2 years ago

Why are the numbers 16 and 21 called relatively prime?

Mathematics

2 answers:

Sergio039 [100]2 years ago

4 0

Answer:

When two numbers have no common factors other than 1.

In other words there is no value that you could divide them both by exactly (without any remainder).

Step-by-step explanation:

labwork [276]2 years ago

3 0

Answer:

Relative prime are the numbers which dont share any of there factors.

Step-by-step explanation:

Relative prime are the numbers which dont share any of there factors. but still they dont share any of there prime cofactors as you can see there factors are different, so they are called as relatively prime. following are the some example of relatively prime numbers . (27,8), (21, 16) , ( 28, 45).

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2 years ago

Select the values that are solutions to the inequality x2 + 3x – 4 > 0.

marusya05 [52]

X ×2 + 3x -4>0
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3 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.

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3 years ago

G (x)=x-3 and h(x)= square root of x, find(g o h) (25)

nevsk [136]

H(25) = sqrt25 = 5

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answer is 2

7 0

2 years ago

How do I find the answer to 10=7-m

andrey2020 [161]

Answer:

M=-3

Step-by-step explanation:

10=7-m

Subtract 7 from both sides

3=-m

Divide by negative

-3=m

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3 years ago