All categories

3 years ago

Who knows how to solve proportions! I NEED HELP!!!!!!! 5/2=2/x x=???

Mathematics

1 answer:

vlada-n [284]3 years ago

3 0

Answer:

$\large\boxed{x=\dfrac{4}{5}=0.8}$

Step-by-step explanation:

$\dfrac{5}{2}=\dfrac{2}{x}\qquad\text{cross multiply}\\\\5x=(2)(2)\\\\5x=4\qquad\text{divide both sides by 5}\\\\\dfrac{5x}{5}=\dfrac{4}{5}\\\\x=\dfrac{4}{5}\to x=0.8$

You might be interested in

Given the following functions:

andrew-mc [135]

Answer:

x^2 - 6x + 9.

Step-by-step explanation:

f(x) = x^2 and g(x) = x - 3.

To find f(g(x)) we replace the x in f(x) by g(x).

f(g(x)) = (x - 3)^2

= x^2 - 6x + 9.

5 0

3 years ago

Find the missing value (x)

Semenov [28]

Answer:

y=14; x=72

Step-by-step explanation:

y=(21×2)/3; x=(48×21)/14

3 0

2 years ago

Find the complex fourth roots of 81(cos(3pi/8) + i sin(3pi/8))

BartSMP [9]

By using De Moivre's theorem:

If we have the complex number ⇒ z = a ( cos θ + i sin θ)
∴ $\sqrt[n]{z} = \sqrt[n]{a} \ (cos \ \frac{\theta + 360K}{n} + i \ sin \ \frac{\theta +360k}{n} )$
k= 0, 1 , 2, ..... , (n-1)

For The given complex number ⇒ z = 81(cos(3π/8) + i sin(3π/8))


Part (A) find the modulus for all of the fourth roots

∴ The modulus of the given complex number = l z l = 81

∴ The modulus of the fourth root = $\sqrt[4]{z} = \sqrt[4]{81} = 3$

Part (b) find the angle for each of the four roots

The angle of the given complex number = $\frac{3 \pi}{8}$
There is four roots and the angle between each root = $\frac{2 \pi}{4} = \frac{\pi}{2}$
The angle of the first root = $\frac{ \frac{3 \pi}{8} }{4} = \frac{3 \pi}{32}$
The angle of the second root = $\frac{3\pi}{32} + \frac{\pi}{2} = \frac{19\pi}{32}$
The angle of the third root = $\frac{19\pi}{32} + \frac{\pi}{2} = \frac{35\pi}{32}$
The angle of the fourth root = $\frac{35\pi}{32} + \frac{\pi}{2} = \frac{51\pi}{32}$

Part (C): find all of the fourth roots of this

The first root = $z_{1} = 3 ( cos \ \frac{3\pi}{32} + i \ sin \ \frac{3\pi}{32})$
The second root = $z_{2} = 3 ( cos \ \frac{19\pi}{32} + i \ sin \ \frac{19\pi}{32})$

The third root = $z_{3} = 3 ( cos \ \frac{35\pi}{32} + i \ sin \ \frac{35\pi}{32})$
The fourth root = $z_{4} = 3 ( cos \ \frac{51\pi}{32} + i \ sin \ \frac{51\pi}{32})$

7 0

3 years ago

HELPPPPPPPPPPPPPPPPPPPP

SVEN [57.7K]

If x + y = 6, then solve for y to get: y = 6 - x.

Now replace y with 6 - x in both equations.

(5x)/3 + 6 - x = c

2(6 - x) = c - 4x

The upper equation is solved for c.
Now we solve the lower equation for c.

c = 2(6 - x) + 4x

c = 12 - 2x + 4x

c = 2x + 12

Since we have two equations solved for c, we substitute to get

(5x)/3 + 6 - x = 2x + 12

This is an equation in only x, so we can solve for x.

(5x)/3 - 3x = 6

5x - 9x = 18

-4x = 18

x = -9/2

Now we solve for y.

x + y = 6

-9/2 + y = 6

y = 9/2 + 12/2

y = 21/2

Now we solve for c.

c = (5x)/3 + y

c = (5 * (-9/2))/3 + 21/2

c = -45/6 + 21/2

c = -15/2 + 21/2

c = 6/2

c = 3

Answer: c = 3

7 0

3 years ago

PLEASE HELP! What is the length of side BC?

yanalaym [24]

The answer is
D 18 3
I'm pretty sure

6 0

3 years ago

Who knows how to solve proportions! I NEED HELP!!!!!!! 5/2=2/x x=???​

Who knows how to solve proportions! I NEED HELP!!!!!!! 5/2=2/x x=???