If someone picks you as the brainiest or wutever, wut does that do?

Solve 5x-4<2x+5. Enter your answer as an inequality.(Need help ASAP pls!!!)

anzhelika [568]

Answer: x < 3

Step-by-step explanation:

To solve this inequality, we will simplify and isolate the x variable.

Given:

5x - 4 < 2x + 5

Add 4 to both sides:

5x < 2x + 9

Subtract 2 from both sides:

3x< + 9

Divide both sides by 3:

x < 3

5 0

2 years ago

Read 2 more answers

Find the cube roots of 27(cos 330° + i sin 330°)

Aleksandr-060686 [28]

Answer:

See below for all the cube roots

Step-by-step explanation:

DeMoivre's Theorem

Let $z=r(cos\theta+isin\theta)$ be a complex number in polar form, where $n$ is an integer and $n\geq1$ . If $z^n=r^n(cos\theta+isin\theta)^n$ , then $z^n=r^n(cos(n\theta)+isin(n\theta))$ .

Nth Root of a Complex Number

If $n$ is any positive integer, the nth roots of $z=rcis\theta$ are given by $\sqrt[n]{rcis\theta}=(rcis\theta)^{\frac{1}{n}}$ where the nth roots are found with the formulas:

$\sqrt[n]{r}\biggr[cis(\frac{\theta+360^\circ k}{n})\biggr]$ for degrees (the one applicable to this problem)
$\sqrt[n]{r}\biggr[cis(\frac{\theta+2\pi k}{n})\biggr]$ for radians

for $k=0,1,2,...\:,n-1$

Calculation

$z=27(cos330^\circ+isin330^\circ)\\\\\sqrt[3]{z} =\sqrt[3]{27(cos330^\circ+isin330^\circ)}\\\\z^{\frac{1}{3}} =(27(cos330^\circ+isin330^\circ))^{\frac{1}{3}}\\\\z^{\frac{1}{3}} =27^{\frac{1}{3}}(cos(\frac{1}{3}\cdot330^\circ)+isin(\frac{1}{3}\cdot330^\circ))\\\\z^{\frac{1}{3}} =3(cos110^\circ+isin110^\circ)$

First cube root where k=2

$\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(2)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+720^\circ}{3})\biggr]\\3\biggr[cis(\frac{1050^\circ}{3})\biggr]\\3\biggr[cis(350^\circ)\biggr]\\3\biggr[cos(350^\circ)+isin(350^\circ)\biggr]$

Second cube root where k=1

$\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(1)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+360^\circ}{3})\biggr]\\3\biggr[cis(\frac{690^\circ}{3})\biggr]\\3\biggr[cis(230^\circ)\biggr]\\3\biggr[cos(230^\circ)+isin(230^\circ)\biggr]$

Third cube root where k=0

$\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(0)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ}{3})\biggr]\\3\biggr[cis(110^\circ)\biggr]\\3\biggr[cos(110^\circ)+isin(110^\circ)\biggr]$

4 0

2 years ago

Will mark you as brainliest for the correct answer!

Alexandra [31]

Answer:

mean= 5+10+15+20+25+30/6=20.82=21

fraction=2+9/24=11/24

3 0

3 years ago

A triangle has a perimeter of 56 cm. Each of the two longer sides of the triangle is three times as long as the shortest side. W

bagirrra123 [75]

The lenght of each side is 24cm, 24cm, and 8cm.

In order to solve this problem, we know that the perimeter of a triangle equation is P = a + b + c, where a, b, and c are the sides of the triangle.

The perimeter is 56cm, we can write the equation as follow:

a + b + c = 56cm (1)

If each of the two longer sides of the triangles is three times as long as the shortest side, we can assume:

c = shortest side = x

a = b = longer sides = 3x

Substituting the values in the equation (1):

3x + 3x + x = 56cm

7x = 56cm

x = 56cm/7 = 8cm

c = shortest side = 8cm

a = b = longer sides = 3(8cm) = 24cm

6 0

3 years ago

an open box is to be made from a piece of metal 16 by 30 inches by cutting out squares of equal size from the corners and bendin

blagie [28]

Let the lengths of the bottom of the box be x and y, and let the length of the squares being cu be z, then
V = xyz . . . (1)
2z + x = 16 => x = 16 - 2z . . . (2)
2z + y = 30 => y = 30 - 2z . . . (3)

Putting (2) and (3) into (1) gives:
V = (16 - 2z)(30 - 2z)z = z(480 - 32z - 60z + 4z^2) = z(480 - 92z + 4z^2) = 480z - 92z^2 + 4z^3
For maximum volume, dV/dz = 0
dV/dz = 480 - 184z + 12z^2 = 0
3z^2 - 46z + 120 = 0
z = 3 1/3 inches

Therefore, for maximum volume, a square of length 3 1/3 (3.33) inches should be cut out from each corner of the cardboard.
The maximum volume is 725 25/27 (725.9) cubic inches.

8 0

3 years ago