1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ludmilkaskok [199]
3 years ago
7

What is the fourth term in this multiplication pattern? multiplication pattern using the formula 2 • 6n – 1 A. 72 B. 12 C. 432 D

. 2592
Mathematics
1 answer:
arlik [135]3 years ago
6 0
4th term = 2 * 6^(4-1) = 2 * 216 = 432 answer
You might be interested in
Help me please.. i don't wanna fail
TEA [102]

Answer:

Step-by-step explanation:

Cindy is correcy because on "Step 1:"  the multiplication was correctly distributed over the subtraction in paranthesis 5(x-2) = 5*x- 5*2 = 5x-10.

Henry was wrrong because he distributed 5 only over x and not over -2 as well 5(x-2) ≠ 5*x-2

5 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
1<br> Find the value of the expression 4a + 7 when a =<br> 2
CaHeK987 [17]

Answer:

15

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Can someone plz help me with this question
Nostrana [21]
10^2 is 10×10 which is 100
4^2 is 4×4 which is 16
so it simplifies to 100-16×2. By Order Of Operations, parentheses, exponents, multiply and divide, add and subtract 100-16×2 makes 100-32 which is 68
5 0
3 years ago
6. Describe how you would make 250 mL of a 4.5 M silver nitrate solution. SHOW ALL CALCS.
Archy [21]

Answer:

Dissolve 190 grams of silver nitrate in 250 mL of water.

Step-by-step explanation:

First, find the moles of AgNO₃ from the volume and concentration.

0.250 L × 4.5 mol/L = 1.125 mol

Next, find the mass of AgNO₃ using the molar mass.

1.125 mol × 170 g/mol = 191.25 g

Rounded to two significant figures, dissolve 190 grams of silver nitrate in 250 mL of water.

8 0
4 years ago
Other questions:
  • Solve 0 Divided by 0
    13·2 answers
  • What is numerical value?
    9·2 answers
  • 1. The sum of any two integers is a whole number.
    13·1 answer
  • Long Division Question of the Day<br> Monday, September 28, 2020<br> 1 5|1989
    7·1 answer
  • Read the excerpt from "An Occurrence at Owl Creek Bridge."
    15·2 answers
  • Can I get get some help please!
    9·1 answer
  • Find the probability of landing in the blue circle on the dart<br> board below.<br> (No links)
    6·1 answer
  • Please help me quickly with this guys it’s 2:43pm and it’s due at 3pm
    13·1 answer
  • Can someone that is good with math help me with this?
    14·1 answer
  • We're planning to enclose a rectangular field with fencing, and we know that for whatever reason we want the field to be 50 feet
    13·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!