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hammer [34]
3 years ago
14

Distance formula for (3,-2) and (5,-3)

Mathematics
1 answer:
Soloha48 [4]3 years ago
7 0

Answer:

d = 2.2

Step-by-step explanation:

Using the distance formula, you would plug in the numbers into the equation. d = √(5 - 3)^2 + ( -3 + 2)^2. So (5 - 3) is 2, and (-3 + 2) is -1. Then square both of the answers, and the equation becomes, d = √4 + 1. And that equals 5, so then you would find the square root, which would be approximately 2.2

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The expression when m=6 and n=4
OverLord2011 [107]
Step 1: plug in the numbers. 7(6)-4
Step 2 multiply 7x6 42-4
7 0
2 years ago
If your dinner bill was $80 and you left a 15% tip. How much money did you leave as a tip? What is your total bill cost?
nadya68 [22]

Answer: 92 dollars because 15% divided into 80 is 12 and add 80 to 12 and its 92

6 0
2 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
2 years ago
Help plz i’ll mark you as brainliest
3241004551 [841]

Answer:what do u need help with?

Step-by-step explanation:

6 0
3 years ago
75=6·w<br> the solution is w=?
Ber [7]
I think answer should be 450 please give me brainlest let me know if it’s correct or not okay thanks bye
7 0
2 years ago
Read 2 more answers
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