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3 years ago

Wich Word means not polite?

Mathematics

2 answers:

gavmur [86]3 years ago

6 0

Answer:

impolite

Step-by-step explanation:

Fofino [41]3 years ago

3 0

Answer :

Impolite or rude, and unmannerly

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Find the slope of the line that passes through the points (1,-4) and (3,-1).

sveta [45]

Answer: E

Step-by-step explanation: I used m.athway

8 0

3 years ago

Write a quadratic equation with -4 and 3/2 as it’s roots. Write the equation in the form ax^2 + bx + c = 0, where a, b, and c ar

jonny [76]

9514 1404 393

Answer:

2x² +5x -12 = 0

Step-by-step explanation:

When p and q are roots of a quadratic, its factored form can be written as ...

(x -p)(x -q) = 0

Here, the roots are given as -4 and 3/2, so the factored form would be ...

(x -(-4))(x -(3/2) = 0

Multiplying by 2 gives us ...

(x +4)(2x -3) = 0

Expanding the product, we find the desired quadratic is ...

2x² +5x -12 = 0

5 0

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