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3 years ago

I need the answer for this, i have posted this multiple times

Mathematics

1 answer:

vladimir1956 [14]3 years ago

7 0

Answer:

1st staement, second option 2nd statement 3rd option, 3rd staement, 4th option, 4th statement, 1st option

Step-by-step explanation:

hope this is right!

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If the ratio of milk cartons to juice boxes is 13:x and there are 39 milk cartons

max2010maxim [7]

The answer is 6 because 39/13=3 and 18/3=6

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2 years ago

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The sum of two consecutive integers is 91. Find the integers.

Reil [10]

Answer:

Let the smaller integer be x and greater integer be (x+1)

x+(x+1)=91

2x+1=91

2x=91-1=90

x=90/2

x=45

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2 years ago

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Lina20 [59]

Answer:

$\textbf{The slope of the parallel line is given by $ 2 $}\\$

Step-by-step explanation:

$\textup{The slope of any two parallel lines are equal and they only differ in their $y- intercepts$. }\\\textup{Given $y = 2x + 4 $}\\\textup{This is comparable to $y = mx + c$, where $m$ is the slope of that line.}\\\textup{Therefore, the slope of the given line and any line parallel to it is $ 2$.}$

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2 years ago

Find the slope of the line that contains the points.<br> (3, 8);(4, -2)

Goshia [24]

Answer:

Is -6

........................

5 0

2 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!

4 0

2 years ago