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3 years ago

Which of the following is most likely the next step in the series? I need help pls :’(

Mathematics

2 answers:

LUCKY_DIMON [66]3 years ago

6 0

Answer:

The answer is: A

Step-by-step explanation:

The way the numbers are counting around the wheel and the color pattern all match A

Leto [7]3 years ago

6 0

Answer:

Step-by-step explanation:

i said so

jk, its cuz the way the numbers count around the wheel make a pattern that matches with A the most

help a sistah, follow their instah: emventures666

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I need to figure out what x is for my homework and I’m so confused teehee.

garik1379 [7]

Answer:

x = 9

Step-by-step explanation:

set a proportion: 8/12 = 6/x => 8x = 72 => x = 9

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3 years ago

After graphing What is the solution of y= -3x -1 and y= x + 3

Tresset [83]

Answer: (3,4)

Step-by-step explanation:

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3 years ago

Jeanne has $12.50 in her bag. How much money does she need to buy a game that costs $72.60?

ololo11 [35]

Answer:

60.1

Step-by-step explanation:

Just get 72.60 and subtract 12.50

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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!

4 0

2 years ago

Order the numbers from least to greatest A.1.81. B.2. C.1.511 D.1.799.

zimovet [89]

B, A, D, C
or 2, 1.81, 1.799, 1.511

3 0

2 years ago

Read 2 more answers