All categories

3 years ago

Need help wit questions 1,2, and 3

Mathematics

1 answer:

12345 [234]3 years ago

7 0

Answer:

number 1 is 6 im not smart tho so you shouldnt put that down

Step-by-step explanation:

You might be interested in

Using the numbers 1 through 9 at most once each time, fill in the blanks to make the equality true

alisha [4.7K]

(9+1-8+2)-(4+5-6+3)=0

6 0

4 years ago

Read 2 more answers

Is the sequence 3, 12, 26..... a geometric sequence? Explain

Thepotemich [5.8K]

$3;\ 12;\ 26;\ ...\\\\If\ a,\ b,\ c\ is\ a\ geometric\ sequence\ then\ ac=b^2.\\\\check:\\3\cdot12=36\neq26$

Answer: The sequence 3, 12, 26, ... is not a geometric sequence.

7 0

3 years ago

Una compañía sabe que si produce "x" unidades mensuales su utilidad "u" se podría calcular con la expresión:

belka [17]

Answer:

Δf(u) /Δx = 92,8 ( razón de cambio promedio)

Step-by-step explanation:

La expresión de la utilidad de la empresa u(x) en función de la cantidad de unidades producidas "x" es:

u(x) = 0,04*x² + 44*x -4000

Entonces la razón de cambio promedio en un intervalo (a ; b) en este caso ( 620 ; 600 ) viene dada por la expresión:

Δf(x)/ Δx = [ f(b) - f(a) ]/( b - a )

en donde f(b) y f(a) se obtienen por sustitución de los valores a y b es decir 600 y 620 respectivamente en la función f(x) = u(x) entonces

Δf(u) /Δx = [ u(b) - u(a) ]/( b -a ) (1)

u(b) = 0,04*(620)² + 44*(620) - 4000

u(b) = 15376 + 27280 -4000

u(b) = 2656 unidades

u(a) = 0,04* (600)² + 44* 600 - 4000

u(a) = 14400 + 26400 - 4000

u(a) = 800

Sustituyendo esos valores en la ecuación 1

Δf(u) /Δx = 2656 - 800 / 620 - 600

Δf(u) /Δx = 92,8

5 0

4 years ago

3) Postcards: Marcie wants to make a box to hold her postcard collection from a piece of

klasskru [66]

The volume of the box is the amount of space in it.

The dimension of the postcard is:

$\mathbf{Length = 10}$

$\mathbf{Width = 18}$

Let x represents the length of the cut.

So, the dimension of the box is:

$\mathbf{Length = 10 -2x}$

$\mathbf{Width = 18 -2x}$

$\mathbf{Height = x}$

(a) The function that represents the volume of the box

This is calculated as:

$\mathbf{V(x) = Length \times Width \times Height}$

So, we have:

$\mathbf{V(x) = (10 - 2x) \times (18 - 2x) \times x}$

Expand

$\mathbf{V(x) = 180x - 56x^2 + 4x^3}$

Hence, the volume function is: $\mathbf{V(x) = 180x - 56x^2 + 4x^3}$

(b) The dimension that maximizes the volume

We have:

$\mathbf{V(x) = 180x - 56x^2 + 4x^3}$

Differentiate

$\mathbf{V'(x) = 180- 112x + 12x^2}$

Set to 0

$\mathbf{180- 112x + 12x^2 = 0}$

Using a calculator, we have:

$\mathbf{x = 7.3,2.1}$

The value x = 7.3 is greater than the dimension of the box.

So, we have:

$\mathbf{x = 2.1}$

Recall that:

$\mathbf{Length = 10 -2x}$

$\mathbf{Width = 18 -2x}$

$\mathbf{Height = x}$

So, we have:

$\mathbf{Length = 10 -2 \times 2.1 = 5.8}$

$\mathbf{Width = 18 -2 \times 2.1 = 13.8}$

$\mathbf{Height = 2.1}$

Hence, the dimension that maximizes the volume is 5.8 by 13.8 by 2.1

(c) The maximum volume

This is calculated as the product of the dimension of the box

So, we have:

$\mathbf{Volume = 5.8 \times 13.8 \times 2.1}$

$\mathbf{Volume = 168.1}$

Hence, the maximum volume of the box is 168.1 cubic inches