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geniusboy [140]
3 years ago
15

Divide 28x3 + 42x2 − 35x by 7x. 4x2 − 6x + 5 4x2 + 6x − 5 4x3 − 6x2 + 5 4x3 + 6x2 − 5

Mathematics
2 answers:
Arturiano [62]3 years ago
6 0
\frac{28x^3+42x^2-35x}{7x}=  \frac{7x(4x^2+6x-5)}{7x}= 4x^2+6x-5
12345 [234]3 years ago
6 0
(28x^3 + 42x^2 - 35x) / 7x

** keep in mind, when dividing exponents, u subtract the exponents

4x^2 + 6x - 5 <==
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Round 2.1 to the nearest whole number.​
Musya8 [376]

Answer:

2

Step-by-step explanation:

if the number after the tenth place is greater than 5 it would be rounded up to 3, but since it is less than 5 it would round to 2

4 0
3 years ago
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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.

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Answer:

Step-by-step explanation:

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I will fan and medal if you help!
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Answer

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Step-by-step explanation:

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