Ask question

Ask question

All categories

3 years ago

14

Can someone please explain why this is incorrect? Isn’t (2x+5)(x+1) the same as (x+1)(2x+5)?

1 answer:

Ne4ueva [31]3 years ago

4 0

Answer:

2(2x+5) (x+1) = 2(x+1)(2x+5)

Step-by-step explanation:

4x^2 +14x+10

Factor out the 2

2(2x^2 +7x+5)

2(2x+5) (x+1)

You need to keep the 2 that you factored out

2(2x+5) (x+1) = 2(x+1)(2x+5) by the Commutative Property of Multiplication

You might be interested in

(x=1)^2-4x=(x-1)^2 simplify and answer

Dennis_Churaev [7]

Step-by-step explanation:

Firstly, there's a typing error at (x=1)². I assume it is (x-1)². If my assumption is correct,then:

x²-2x+1-4x=x²-2x+1

x²-6x+1-x²+2x-1=0

-4x=0

x=0

5 0

2 years ago

Arrange these fractions in least to greatest order:

earnstyle [38]

1 6/7 ................

7 0

2 years 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

Mrs. Gordon bought 10 kg of apples for $16.50. What is the price per pound of apples?

natulia [17]

Answer:

1.65 per kg

Step-by-step explanation:

4 0

3 years ago

Match the verbal expression (term) with its algebraic expression (definition).

serious [3.7K]

Answer:

match

Step-by-step explanation:

4 0

3 years ago