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lubasha [3.4K]
3 years ago
12

HELP PLEASE ASAP!!!!!!!!!!

Mathematics
2 answers:
Mazyrski [523]3 years ago
8 0

Answer:

A

Step-by-step explanation:

laila [671]3 years ago
5 0

Answer:

B

Also:

mark me brainliest if this helped plz :)

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3. List the coefficients in the expression below. 14x - 7x + 1-2x​
elena-14-01-66 [18.8K]

Answer:

14, -7, -2

Step-by-step explanation:

<u>The coefficient is the number that multiplies the variable, so list the numbers that fall under that definition:</u>

14x - 7x + 1 - 2x

↑      ↑          ↑

14    -7         -2

<u>The coefficients in the expression are </u><u>14, -7, and -2.</u>

7 0
2 years ago
Use the zeros of the following quadratic to find the x-value of the vertex.
frez [133]
(x-1)^2 is the factored version, so vertex is at 0, which is 1, so, A
6 0
3 years ago
У + 2х = 7<br> у = 3х – 3<br> What is the answer to this
Neko [114]
I think the correct answer is ( 2, 3 )
6 0
3 years ago
Find each probability below -- as a fraction, decimal and percent <br> Will give brainliest
Komok [63]

Answer:  

I was not sure to what decimal point to round it to

a) Fraction: 2/7    Decimal: 0.28571   Percent: 28.571 %

b) Fraction: 5/7    Decimal: 0.71428   Percent: 71.428 %

c) Fraction: 1/7     Decimal: 0.14285   Percent: 14.285 %

7 0
3 years ago
Read 2 more answers
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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