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3 years ago

6

If f(x) = 2(x − 5), find f(8).

1 answer:

Jet001 [13]3 years ago

4 0

Answer:

6

Step-by-step explanation:

f(x) = 2(x-5)

f(8) = 2(8-5)

parentheses comes first so, 8-5 = 3
2(3), so multiply and the final answer is 6

f(8) = 6

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Please answer the question in the picture below. Thanks!

Arada [10]

Answer:

Yes

Step-by-step explanation:

$\frac{DH}{HE}=\frac{18}{36}=\frac{1}{2}$

$\text{This means that } DH=\frac{1}{2}HE$

and thus the given segments are parallel by the converse of the triangle proportionality theorem.

4 0

1 year ago

Find the mass of the lamina that occupies the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with the density function ρ(x, y) = xye

Alona [7]

Answer:

The mass of the lamina is 1

Step-by-step explanation:

Let $\rho(x,y)$ be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:

$m=\int\limits \int\limits_D \rho(x,y) \, dA$ .

From the question, the given density function is $\rho (x,y)=xye^{x+y}$ .

Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

The mass of the lamina can be found by evaluating the double integral:

$I=\int\limits^1_0\int\limits^1_0xye^{x+y}dydx$ .

Since D is a rectangular region, we can apply Fubini's Theorem to get:

$I=\int\limits^1_0(\int\limits^1_0xye^{x+y}dy)dx$ .

Let the inner integral be: $I_0=\int\limits^1_0xye^{x+y}dy$ , then

$I=\int\limits^1_0(I_0)dx$ .

The inner integral is evaluated using integration by parts.

Let $u=xy$ , the partial derivative of u wrt y is

$\implies du=xdy$

and

$dv=\int\limits e^{x+y} dy$ , integrating wrt y, we obtain

$v=\int\limits e^{x+y}$

Recall the integration by parts formula: $\int\limits udv=uv- \int\limits vdu$

This implies that:

$\int\limits xye^{x+y}dy=xye^{x+y}-\int\limits e^{x+y}\cdot xdy$

$\int\limits xye^{x+y}dy=xye^{x+y}-xe^{x+y}$

$I_0=\int\limits^1_0 xye^{x+y}dy$

We substitute the limits of integration and evaluate to get:

$I_0=xe^x$

This implies that:

$I=\int\limits^1_0(xe^x)dx$ .

Or

$I=\int\limits^1_0xe^xdx$ .

We again apply integration by parts formula to get:

$\int\limits xe^xdx=e^x(x-1)$ .

$I=\int\limits^1_0xe^xdx=e^1(1-1)-e^0(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(-1)=1$ .

No unit is given, therefore the mass of the lamina is 1.

3 0

3 years ago

A rectangle is 1/2 ft long and 3/4 ft wide.

Alex17521 [72]

Area = Length * Width
A = 1/2 * 3/4
A = 3/8

In short, Your Answer would be 3/8 Ft^2

Hope this helps!

8 0

2 years ago

Read 2 more answers

Let W (a) = a^3- a^2 Find W (2)

grin007 [14]

Answer:

4

Step-by-step explanation:

plug in 2 for the a variables and you get 4 as the answer.

8 0

2 years ago

Please help 50 points!<br> Simplify the expressed

LenKa [72]

Answer:

Step-by-step explanation:

$\frac{x^{2} -x-12}{x+4} * \frac{2x^{2}-32 }{x^{2} -8x+16}$ rewrite as multiplication

$\frac{x^{2} +3x-4x-12}{x+4} * \frac{2(x^{2}-16) }{x^{2} -8x+16}$ look to factor

$\frac{x(x+3)-4(x+3)}{x+4} * \frac{2(x-4)(x+4) }{(x-4)^{2}}$ , difference of 2 squares and perfect square

$\frac{(x+3)(x-4)}{x+4} * \frac{2(x-4)(x+4) }{(x-4)^{2}}$ , finish factoring and simplify

2(x+3)

3 0

2 years ago