All categories

2 years ago

Given f(x)=5x+15 find f(x)=30

1 answer:

3 0

Given the function, f(x) = 5x + 15:

In order to find the value of x when f(x) = 30, substitute the value of the output into the function:

f(x) = 5x + 15
30 = 5x + 15

Subtract 15 from both sides:
30 - 15 = 5x + 15 - 15

15 = 5x

Divide both sides by 5:
15/5 = 5x/5

x = 3

Therefore, the value of x when f(x) = 30 is 3, or f(3) = 30.

You might be interested in

What is the domain of the square root function graphed below

Lina20 [59]

Answer:

Dominio de una Raíz cuadrada

Cuando tenemos una raíz, para estudiar su existencia debemos saber que una raíz existe si y solo sí el radicando (lo que está dentro de la raíz) sea mayor o igual a cero.

Por lo tanto deberemos estudiar el signo de lo que se encuentra dentro de la raíz; y determinar las zonas positivas y las raíces que son las zona de existencia de la función raíz o dominio de esta función.

Step-by-step explanation:

qsy

5 0

2 years ago

Six equilateral triangles are connected to create a regular hexagon. The area of the hexagon is 24a2 – 18 square units. Which is

liberstina [14]

<span>Six equilateral triangles are connected to create a regular hexagon. The area of the hexagon is 24a2 – 18 square units. Which is an equivalent expression for the area of the hexagon based on the...

Answer: </span> 6(4a2 - 3)

hope this helps :)

8 0

2 years ago

Read 2 more answers

Drone INC. owns four 3D printers (D1, D2, D3, D4) that print all their Drone parts. Sometimes errors in printing occur. We know

USPshnik [31]

Answer:

Step-by-step explanation:

Hello!

There are 4 3D printers available to print drone parts, then be "Di" the event that the printer i printed the drone part (∀ i= 1,2,3,4), and the probability of a randomly selected par being print by one of them is:

D1 ⇒ P(D1)= 0.15

D2 ⇒ P(D2)= 0.25

D3 ⇒ P(D3)= 0.40

D4 ⇒ P(D4)= 0.20

Additionally, there is a chance that these printers will print defective parts. Be "Ei" represent the error rate of each print (∀ i= 1,2,3,4) then:

P(E1)= 0.01

P(E2)= 0.02

P(E3)= 0.01

P(E4)= 0.02

Ei is then the event that "the piece was printed by Di" and "the piece is defective".

You need to determine the probability of randomly selecting a defective part printed by each one of these printers, i.e. you need to find the probability of the part being printed by the i printer given that is defective, symbolically: P(DiIE)

Where "E" represents the event "the piece is defective" and its probability represents the total error rate of the production line:

P(E)= P(E1)+P(E2)+P(E3)+P(E4)= 0.01+0.02+0.01+0.02= 0.06

This is a conditional probability and you can calculate it as:

$P(A/B)= \frac{P(AnB)}{P(B)}$

To reach the asked probability, first, you need to calculate the probability of the intersection between the two events, that is, the probability of the piece being printed by the Di printer and being defective Ei.

P(D1∩E)= P(E1)= 0.01

P(D2∩E)= P(E2)= 0.02

P(D3∩E)= P(E3)= 0.01

P(D4∩E)= P(E4)= 0.02

Now you can calculate the probability of the piece bein printed by each printer given that it is defective:

$P(D1/E)= \frac{P(E1)}{P(E)} = \frac{0.01}{0.06}= 0.17$

$P(D2/E)= \frac{P(E2)}{P(E)} = \frac{0.02}{0.06}= 0.33$

$P(D3/E)= \frac{P(E3)}{P(E)} = \frac{0.01}{0.06}= 0.17$

$P(D4/E)= \frac{P(E4)}{P(E)} = \frac{0.02}{0.06}= 0.33$

P(D2)= 0.25 and P(D2/E)= 0.33 ⇒ The prior probability of D2 is smaller than the posterior probability.

The fact that P(D2) ≠ P(D2/E) means that both events are nor independent and the occurrence of the piece bein defective modifies the probability of it being printed by the second printer (D2)

I hope this helps!

8 0

2 years ago

The perimeter of the small square is equal to one-half the perimeter of the large square,

Vsevolod [243]

I hope this helps you