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2 years ago

Does xxxxx = 5x? I'm not really sure so it would help.

Mathematics

1 answer:

hodyreva [135]2 years ago

6 0

Answer:

Step-by-step explanation:

no ,x*x*x*x*x=x^5

x+x+x+x+x=5x

in multiplication those who have same base we add their power.

in addition those variables which have same base(or are like term) we add them.

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Write an equation of the line that passes through point (-8, – 5) and has the slope m= - 1/4 y=

zmey [24]

Answer:

y= -1/4x + 3

Step-by-step explanation:

y-y1 = m(x-x1)

(-8 {x1}, -5 {y1}, -1/4 [m] )

We'll then replace the letters with the available figures.

y-[-5] = -1/4 (x-[-8])

We are made to know that minus + minus= plus

Minus + plus= Minus

Minus x minus = plus

minus x plus = minus

Therefore, the mathematical setting of this question becomes,

y+5 = -1/4 (x+8)

Multiply everything in the bracket with the outside value

-1/4 x X = -1/4x

-1/4 x 8 = -2

The mathematical setting then becomes,

y+5 = -1/4x + -2 (Note: plus + minus = minus)

so, it becomes

y+5 = -1/4x - 2

The equation is supposed to look like this, y= mx + b

Therefore we add +5 to both sides to cancel out

y+5 = -1/4x - 2

+5 +5

+5 cancels out +5 which makes the setting,

y= -1/4x - 2 +5 (-2+5 goes around to become 5-2= 3)

The expression now looks like this

y= -1/4x +3 which is the final answer!

$y= -1/4x + 3$

5 0

1 year 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

3 years ago

Can someone please help me find the simple interest

Stells [14]

$\bf ~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$300\\ r=rate\to 4\%\to \frac{4}{100}\dotfill &0.04\\ t=years\dotfill &3 \end{cases} \\\\\\ I=(300)(0.04)(3)\implies I=36$

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3 years ago

Helpppppp me please

motikmotik

I believe it’s 100 hopefully it helps

6 0

3 years ago

Solve the system by substitution. y = 7x – 20 y = 2x

g100num [7]

Answer:

x=4 y=8

Step-by-step explanation:

Solve for the first variable in one of the equations, then substitute the result into the other equation

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2 years ago

Does x*x*x*x*x = 5x? I'm not really sure so it would help.

Does xxxxx = 5x? I'm not really sure so it would help.