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3 years ago

How do you find the area of 7 & 8?

Mathematics

2 answers:

saveliy_v [14]3 years ago

8 0

Area= Length x Width
7: 48m 2
8: 78.37375 ft 2

Wittaler [7]3 years ago

3 0

For the first one split into 3. Find the area of the 2 triangles by timesing base by height and add the square in the middle. Sorry if this didn't make sende

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A bag contains:

Lady bird [3.3K]

Answer:

3/50

Step-by-step explanation:

There are 5 + 4 + 2 + 6 + 3 = 20 marbles in the bag

P(draw a black) = 6/20 = 3/10

If the black marble is replaced, then there are 20 marbles in the bag again.

P(draw a blue) = 4/20 = 1/5

P(draw a black and blue with replacement) = (3/10)(1/5) = 3/50

6 0

3 years ago

Simplify.

Hoochie [10]

C might be the answer

7 0

3 years ago

Use the x-intercept method to find all real solutions of the equation. x^3-9x^2+11x+21=0

IgorLugansk [536]

We must graph y=x^3-9x^2+11x+21. Please, see the attached file.
Then we see were the graph cuts the x-axis, and these values of x are the real solutions of the given equation:

The real solutions of the given equation are:x = -1, 3, and 7

7 0

3 years ago

Y=x^2-25 is it linear?

Svetllana [295]

Answer:

yes it is i think

7 0

2 years ago

Read 2 more answers

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