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3 years ago

-3x-4(x-1)=5(2x+1)+6x

Mathematics

1 answer:

Firdavs [7]3 years ago

6 0

Answer:

x = (-1)/23

Step-by-step explanation:

Solve for x:

-3 x - 4 (x - 1) = 5 (2 x + 1) + 6 x

-4 (x - 1) = 4 - 4 x:

4 - 4 x - 3 x = 5 (2 x + 1) + 6 x

Grouping like terms, -3 x - 4 x + 4 = 4 + (-3 x - 4 x):

4 + (-3 x - 4 x) = 5 (2 x + 1) + 6 x

-3 x - 4 x = -7 x:

-7 x + 4 = 5 (2 x + 1) + 6 x

5 (2 x + 1) = 10 x + 5:

4 - 7 x = 10 x + 5 + 6 x

Grouping like terms, 10 x + 6 x + 5 = (10 x + 6 x) + 5:

4 - 7 x = (10 x + 6 x) + 5

10 x + 6 x = 16 x:

4 - 7 x = 16 x + 5

Subtract 16 x from both sides:

4 + (-7 x - 16 x) = (16 x - 16 x) + 5

-7 x - 16 x = -23 x:

-23 x + 4 = (16 x - 16 x) + 5

16 x - 16 x = 0:

4 - 23 x = 5

Subtract 4 from both sides:

(4 - 4) - 23 x = 5 - 4

4 - 4 = 0:

-23 x = 5 - 4

5 - 4 = 1:

-23 x = 1

Divide both sides by -23:

Answer: x = (-1)/23

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3 years ago

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The slope-intercept form of a line:

$y=mx+b\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\to slope\\\\b\to y-intercept$

We have:

$(-3,\ 2)\to x_1=-3,\ y_1=2\\(1,\ 5)\to x_2=1,\ y_2=5$

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Put the values of coordinates of the point (1, 5) to the equation of a line:

$5=\dfrac{3}{4}(1)+b\\\\5=\dfrac{3}{4}+b\qquad|-\dfrac{3}{4}\\\\b=4\dfrac{1}{4}$

Answer: $y=\dfrac{3}{4}x+4\dfrac{1}{4}$

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3 years ago

A flat circular plate has the shape of the region x squared plus y squared less than or equals 1.The plate, including the bound

rjkz [21]

Answer:

We have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . and the hottest value is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

Step-by-step explanation:

We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.

The temperature equation is:

$T(x,y)=x^{2}+2y^{2}-\frac{3}{2}x$

Let's take the partials derivatives.

$T_{x}(x,y)=2x-\frac{3}{2}=0$

$T_{y}(x,y)=4y=0$

So, we can find the critical point (x,y) of T(x,y).

$2x-\frac{3}{2}=0$

$x=\frac{3}{4}$

$4y=0$

$y=0$

The critical point is (3/4,0) so the temperature at this point is: $T(\frac{3}{4},0)=(\frac{3}{4})^{2}+2(0)^{2}-(\frac{3}{2})(\frac{3}{4})$

$T(\frac{3}{4},0)=-\frac{9}{16}$

Now, we need to evaluate the boundary condition.

$x^{2}+y^{2}=1$

We can solve this equation for y and evaluate this value in the temperature.

$y=\pm \sqrt{1-x^{2}}$

$T(x,\sqrt{1-x^{2}})=x^{2}+2(1-x^{2})-\frac{3}{2}x$

$T(x,\sqrt{1-x^{2}})=-x^{2}-\frac{3}{2}x+2$

Now, let's find the critical point again, as we did above.

$T_{x}(x,\sqrt{1-x^{2}})=-2x-\frac{3}{2}=0$

$x=-\frac{3}{4}$

Evaluating T(x,y) at this point, we have:

$T(-(3/4),\sqrt{1-(-3/4)^{2}})=-(-\frac{3}{4})^{2}-\frac{3}{2}(-\frac{3}{4})+2$

$T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$

Now, we can see that at point (3/4,0) we have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . On the other hand, at the point $-(3/4),\frac{\sqrt{7}}{4})$ we have the hottest value of temperature, it is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

I hope it helps you!

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2 years ago