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3 years ago

Find the value of “x” so that a parallel to b

Mathematics

2 answers:

Kaylis [27]3 years ago

8 0

Answer:

Step-by-step explanation:

3x - 50 = 2x - 5 { a// b, so alternate interior angles are equal}

3x - 2x = -5 +50

x = 45

Doss [256]3 years ago

5 0

X = 45 ............

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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:

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

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

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 medical clinic is reducing the number of incoming patients by giving vaccines before flu season. During week 5 of flu season,

8_murik_8 [283]

Answer:

$f(x) = -6x +120$

Step-by-step explanation:

Let's call y the number of patients treated each week

Let's call x the week number.

If the reduction in the number of patients each week is linear then the equation that models this situation will have the following form:

$y = mx + b$

Where m is the slope of the equation and b is the intercept with the x-axis.

If we know two points on the line then we can find the values of m and b.

We know that During week 5 of flu season, the clinic saw 90 patients, then we have the point:

(5, 90)

We know that In week 10 of flu season, the clinic saw 60 patients, then we have the point:

(10, 60).

Then we can find m and b using the followings formulas:

$m=\frac{y_2-y_1}{x_2-x_1}$ and $b=y_1-mx_1$

In this case: $(x_1, y_1) = (5, 90)$ and $(x_2, y_2) = (10, 60)$

Then:

$m=\frac{60-90}{10-5}$

$m=-6$

And

$b=90-(-6)(5)$

$b=120$

Finally the function that shows the number of patients seen each week at the clinic is:

$f(x) = -6x +120$

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evablogger [386]

The formula for the perimeter of a rectangle is P = 2L + 2W, where L is the length and W is the width. Because we don't know either the length or the width we can't solve the problem...too many unknowns. BUT we do have some information that will help with this problem. We are told that the length is 2 feet longer than the width, so we can use that: L = W+2. Now we can make the substitution into the formula along with the value for the perimeter that was given to us: 36=2(W+2) + 2W, and 36 = 2W + 4 + 2W; 36 = 4W + 4; 32 = 4W and W = 8. Now go back to where you said that the length is 2 feet longer than the width. If the width is 8, then 8+2 = 10 for the length.

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3 years ago

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Nezavi [6.7K]

Answer:

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Step-by-step explanation:

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