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iragen [17]
3 years ago
10

To dived 572÷4, Stanley estimated to place the first digit of the quotient. In which place is the first digit of the question?

Mathematics
1 answer:
Helga [31]3 years ago
5 0

Answer:


Step-by-step explanation:

i am not do sure but i think its in the ones place...

-sorry if im wrong


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Prove that for any natural n the number 3^(n+4)−3n is divisible by 16.
Pani-rosa [81]

Answer:

Step-by-step explanation:

Prove that for any natural number 3^(n+4)-3n is divisible by 16.

(I'm going to assume that you mean 3^(n+4)-3^n.)

1. We can break up 3^(n+4)-3n into 3^n * 3^4-3^n (by the rule a^b*a^c = a^b+c).

2. Solve to get 3^n * 81 - 3^n

3. Factor out the 3^n, and you'll get 3^n(81-1), and simplify: 3^n(80)

        You may notice that 80 is divisible by 16.

4. Rewrite what we got from the last step as: 3^n*5(16).

Hope this helped you!

3 0
3 years ago
Suppose that f(x) = x2 and g(x) = x2. Which statement best compares the graph of g(x) with the graph of f(x)?
Nesterboy [21]
The anaswer would be a because if it is streched that long then the graph would be 2

5 0
3 years ago
Read 2 more answers
Size Cost 4R = $0.25 per photo 5R = $0.50 Per photo 6R = $0.80 per photo A school ordered 308 photos of size 4R and some photos
lbvjy [14]

Answer:

0.25×308=$77

185-77=$108

108/0.50×1

=216photos

Total photos=216+308

=524photos

Step-by-step explanation:

3 0
3 years ago
Write an equation for the parabola whose vertex is at (4, 12) and which passes through (5, 21)
laiz [17]

Answer:

y - 12 = 9(x - 4)

Step-by-step explanation:

The vertex (h, k) is (4, 12) and the point (5, 21) is on the graph.  Assuming that this is a vertical parabola, opening up (because the coordinate 21 is greater than the coordinate 12), we insert the knowns into  y - k = a(x - h)^2, obtaining

21 - 12 = a(5 - 4), or  9 = a.  With a known, we can write the desired equation:

y - 12 = 9(x - 4)

8 0
3 years ago
Find the critical points of the surface f(x, y) = x3 - 6xy + y3 and determine their nature.​
Vedmedyk [2.9K]

Compute the gradient of f.

\nabla f(x,y) = \left\langle 3x^2 - 6y, -6x + 3y^2\right\rangle

Set this equal to the zero vector and solve for the critical points.

3x^2-6y = 0 \implies x^2 = 2y

-6x+3y^2=0 \implies y^2 = 2x \implies y = \pm\sqrt{2x}

\implies x^2 = \pm2\sqrt{2x}

\implies x^4 = 8x

\implies x^4 - 8x = 0

\implies x (x-2) (x^2 + 2x + 4) = 0

\implies x = 0 \text{ or } x-2 = 0 \text{ or } x^2 + 2x + 4 = 0

\implies x = 0 \text{ or } x = 2 \text{ or } (x+1)^2 + 3 = 0

The last case has no real solution, so we can ignore it.

Now,

x=0 \implies 0^2 = 2y \implies y=0

x=2 \implies 2^2 = 2y \implies y=2

so we have two critical points (0, 0) and (2, 2).

Compute the Hessian matrix (i.e. Jacobian of the gradient).

H(x,y) = \begin{bmatrix} 6x & -6 \\ -6 & 6y \end{bmatrix}

Check the sign of the determinant of the Hessian at each of the critical points.

\det H(0,0) = \begin{vmatrix} 0 & -6 \\ -6 & 0 \end{vmatrix} = -36 < 0

which indicates a saddle point at (0, 0);

\det H(2,2) = \begin{vmatrix} 12 & -6 \\ -6 & 12 \end{vmatrix} = 108 > 0

We also have f_{xx}(2,2) = 12 > 0, which together indicate a local minimum at (2, 2).

3 0
2 years ago
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