All categories

3 years ago

Name the property of real numbers illustrated by the equation 16(3t + 4v) = 48t + 64v.

1 answer:

6 0

The property of real numbers illustrated by the equation 16(3t+4v)=48t+ 64v.

Therefore, the answer is: Distributive property.

As you can see, you have the equation 16(3t+4v), and when you apply the Distributive property, you obtain:

16(3t+4v)
(16)(3t)+(16)(4v)
48t+64v

You might be interested in

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

The branch manager of an outlet (Store 1) of a nationwide

monitta

Answer:

1. ( 19.1416, 23.5384,)

2. (0.276348, 0.46651)

3. the sample size = 170.73 approximately 171

4. sample size = 334.07 approximately 334

5. sample of 334 should be taken by manager

Step-by-step explanation:

mean = bar x = 21.34 dollars

size of sample n = 70

standard deviation of sample = 9.22

we use t distribution as the population standard deviation is not known.

95% Confidence interval

1-α = 0.95

α = 0.05

degree of freedom = 70-1 = 69

α/2 = 0.025

using the t distribution tsble,

= 1.9949

confidence interval = $21.34+-1.9949*[\frac{9.22}{sqrt(70)}] \\$

= 21.34 +- (1.9949*1.10200)

= 21.34 + 2.1984, 21.34 - 2.1984

= (23.5384, 19.1416)

the confidence interval of the mean amount spent at the supply store can be written as 19.1416<u<23.5384

2. sixe of those who only have a cat

p = 26/70 = 0.371429

at 90 % confidence interval,

1-α = 0.90

α = 0.10

we use the z table here

z(0.10/2) = Z(0.05)

= 1.645

$0.371429+-1.645\sqrt} \frac{0.371429(1-0.371429)}{70}$

= 0.371429 +-( 1.645 x 0.0578)

= 0.371429 + 0.095081, 0.371429 - 0.095081

= (0.276348, 0.46651)

3. sd = 10$

margin of erro,r e = 1.50$

α = 0.05

using z table

α/2 = Z0.025

= 1.96

sample size = 1.96² * 10² / 1.50²

= 3.8416 * 100/ 2.25

= 170.73

the sample size is approximately 171

d. we have 0.5 as sample proportion now

margin of error = 0.045

α = 0.10

Zα/2 = 0.05

= 1.645

sample size = 1.645²x0.5(1-0.5) / 0.045²

= 0.676506/0.002025

= 334. 07

sample size = 334

5. sample of 334 should be taken by manager

3 0

3 years ago

Shakira went bowling with her friends. She paid $3 to rent shoes and then $4.75 for each game of bowling. If she spent a total o

DaniilM [7]

They played 9.17 games

6 0

3 years ago

Read 2 more answers

Simplify: 6-4[3(2x+5)-4x]

Degger [83]

[- 12 - 8x +5 16x]+ 6

7 0

3 years ago

Read 2 more answers

A triangle ABC has its vertices at A(-2, -3), B(2, 1), and C(5,-1).

maksim [4K]

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill ~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-3})\qquad B(\stackrel{x_2}{2}~,~\stackrel{y_2}{1}) ~\hfill AB=\sqrt{( 2- (-2))^2 + ( 1- (-3))^2} \\\\\\ AB=\sqrt{(2+2)^2+(1+3)^2}\implies AB=\sqrt{32}\implies \boxed{AB=4\sqrt{2}} \\\\[-0.35em] ~\dotfill$

$B(\stackrel{x_1}{2}~,~\stackrel{y_1}{1})\qquad C(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) ~\hfill BC=\sqrt{( 5- 2)^2 + ( -1- 1)^2} \\\\\\ BC=\sqrt{3^2+(-2)^2}\implies BC=\sqrt{9+4}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ C(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad A(\stackrel{x_2}{-2}~,~\stackrel{y_2}{-3}) ~\hfill CA=\sqrt{( -2- 5)^2 + ( -3- (-1))^2} \\\\\\ CA=\sqrt{(-7)^2+(-3+1)^2}\implies CA=\sqrt{49+(-2)^2}\implies \boxed{CA=\sqrt{53}}$

$\stackrel{\textit{\large perimeter of ABC}}{4\sqrt{2}+\sqrt{13}+\sqrt{53}~~\approx~~ 16.54}$

3 0

2 years ago