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3 years ago

11. The real number

Mathematics

1 answer:

victus00 [196]3 years ago

5 0

Answer:

B whole number

Step-by-step explanation:

Whole number is the right answer because integers are even numbers, irrational numbers are in root. and rational numbers are in points.

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Please Help Me With This.

Romashka-Z-Leto [24]

Answer:

A is $0.45 B is 2 ounces

Step-by-step explanation:

A when you divide the cost by the ounces or $5.40 by 12 ounces you get $0.45 per ounce.

B $0.45 times 2 equals 90 cents and you cannot go over 1 dollar.

C you should be to do along with D with this information.

5 0

3 years ago

Write the decimal expansion for 7/9

dedylja [7]

0.7777777778
there you go your welcome

6 0

3 years ago

Read 2 more answers

3) There were 35 children and 10 adults at a barbeque.

Mila [183]

Answer:

3. a. 10:35= 2:7

b.35:45=7:9

c.40:50=4:5

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

I hope it helps you!

4 0

2 years ago

A runner starts from rest and accelerates uniformly to a speed of 8.0 meters per

ivanzaharov [21]

2 m/s²

Step-by-step explanation:

Step 1:

Let the initial velocity be 0 m/s and the final velocity be 8 m/s and the time traveled is 4 s.

From the basic science,

Acceleration = ( Final Velocity - Initial Velocity) / Time

Step 2:

Acceleration = (8-0)/4

2 m/s²

4 0

3 years ago