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Sedbober [7]
3 years ago
6

Which expression is equivalent to (3m)(4m)^2

Mathematics
2 answers:
GenaCL600 [577]3 years ago
5 0

(3x) \times (4x) {}^{2}  \\ 3x \times 16x {}^{2}  \\ 48x {}^{3}

topjm [15]3 years ago
4 0

Answer:

C. 48m^3

Step-by-step explanation:

Edge2020 bih

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The function g(x) = x2 is transformed to obtain function h:
Rudik [331]

Answer:

Option C

The graph of g is vertically shifted 5 units down

4 0
3 years ago
Solve the following system of equations by substitution<br> 2x – 3y = -1<br> y = x - 1
Aliun [14]

Answer:

(4, 3)

Step-by-step explanation:

Plug in x - 1 where y is. The equation will turn in 2x - 3(x - 1) = -1. Distribute, and you'll have 2x -3x + 3 = -1. Combine like terms (2x and -3x) and you'll have -x + 3 = -1. Subtract 3 from both sides, and you'll have -x = -4. Divide by -x (or -1) on both sides, and you'll have x = 4. Since y = x - 1, y = 3, since 4 - 1 = 3. Sorry if this sounds repetitive btw. Good luck :D

7 0
3 years ago
ABCD is a parallelogram. Find the indicated measure:
uranmaximum [27]

Answer:

12 12 12 12 12 12 12 12 12 1 2 12 12

7 0
3 years ago
Solve for x in the triangle above
faust18 [17]

Answer:

x = 30

Step-by-step explanation:

The interior angles of a triangle must always add up to 180.

2x + 2x + 3x - 30 = 180

Add 30 to both sides to seperate variables.

2x + 2x + 3x = 210

Combine like terms.

7x = 210

Divide by 7 to isolate x.

x = 30

8 0
3 years ago
Show work please.<br><br> solve system of equations using matrices.
nadya68 [22]

Answer:

(t, t -1, t)

Step-by-step explanation:

You have three unknowns but only 2 equations, so you can't really SOLVE this...you can get a solution with a variable still in it (I forget what this is called.  I think it refers to infinite many solutions).  Here's how it works:

Set up your matrix:

\left[\begin{array}{ccc}1&-2&1\\2&-1&-1\\\end{array}\right] \left[\begin{array}{ccc}2\\1\\\end{array}\right]

You want to change the number in position 21 (the 2 in the scond row) to a 0 so you have y and z left.  Do this by multiplying the top row by -2 then adding it to the second row to get that 2 to become a 0.  Multiplying in a -2 to the top row gives you:

\left[\begin{array}{ccc}-2&4&-2\\2&-1&-1\\\end{array}\right]\left[\begin{array}{ccc}-4\\1\\\end{array}\right]

Then add, keeping the first row the same and changing the second to reflect the addition:

\left[\begin{array}{ccc}-2&4&-2\\0&3&-3\\\end{array}\right] \left[\begin{array}{ccc}-4\\-3\\\end{array}\right]

The second equation is this now:

3y - 3z = -3.  Solving for y gives you y = z - 1.  Let's let z = t (some random real number that will make the system true.  Any number will work.  I'll show you at the end.  Just bear with me...)

lf z = t, and if y = z - 1, then y = t - 1.  So far we have that y = t - 1 and z = t.  Now we solve for x:

From the first equation in the original system,

x - 2y + z = 2.  Subbing in t - 1 for y and t for z:

x - 2(t - 1) + t = 2.  Simplify to get

x - 2t + 2 + t = 2  and  x - t = 0, and x = t.  So the solution set is (t, t - 1, t).  Picking a random value for t of, let's say 2, sub that in and make sure it works.  If:

x - 2y + z = 2, then t - 2(t - 1) + t = 2 becomes t - 2t + 2 + t = 2, and with t = 2, 2 - 2(2) + 2 + 2 = 2.    Check it:  2 - 4 + 4 = 2 and 2 = 2.  You could pick any value for t and it will work.

6 0
3 years ago
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