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lord [1]
3 years ago
14

Point A is located at (1, 5), and point A′ is located at (-2, 1). What is the distance from B to B′ ?

Mathematics
1 answer:
Lunna [17]3 years ago
5 0

Answer:

leonelbhsusnsbsoaosbh

Step-by-step explanation:

osbdhdudjhduxhdhsususisbejoebeiwhwblepb2093764

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Please help. I'm stuck.
Bezzdna [24]
The root \sqrt{10} can be converted into the power 10^{ \frac{1}{2} }. Therefore we can rewrite the problem as (10^{ \frac{1}{2} } )^ {\frac{3}{4} x} and then follow the exponent rules about a power to a power, multiplying 1/2 and 3/4 together.

Thus the problem becomes 10^{\frac{3}{8}x}, which then can be turned into \sqrt[8]{10} ^{3x}, making the last option our answer.

4 0
3 years ago
-5x+3y=6 in slope intercept form
Leno4ka [110]

slope intercept form is y=mx+b

To find this from standard form you need to remove x to the other side. To do this you need to add 5x to both sides. You will end up with 3y=5x+6. To yet the y by itself you need to divide everything in the equation by 3. You would end up with y=5/3x +2 which is the equation in slope intercept form

8 0
3 years ago
How do you compare 492,111 and 409,867
ExtremeBDS [4]
By adding an < or >.
6 0
3 years ago
Read 2 more answers
Use the graph to write a linear function that relates y to x.
allsm [11]

Answer:

um we need the photo sry

Step-by-step explanation:

6 0
3 years ago
I AM GIVING 45 POINTS TO WHOEVER GETS THIS RIGHT... plz answer corectly and try... i was working on this for sooo long. Plz try
Alexxx [7]
When you have 3 choices for each of 6 spins, the number of possible "words" is
  3^6 = 729

The number of permutations of 6 things that are 3 groups of 2 is
  6!/(2!×2!×2!) = 720/8 = 90

A) The probability of a word containing two of each of the letters is 90/729 = 10/81


The number of permutations of 6 things from two groups of different sizes is
  (2 and 4) : 6!/(2!×4!) = 15
  (3 and 3) : 6!/(3!×3!) = 20
  (4 and 2) : 15
  (5 and 1) : 6
  (6 and 0) : 1

B) The number of ways there can be at least 2 "a"s and no "b"s is
  15 + 20 + 15 + 6 + 1 = 57
The probability of a word containing at least 2 "a"s and no "b"s is 57/729 = 19/243.


_____
These numbers were verified by listing all possibilities and actually counting the ones that met your requirements.
6 0
3 years ago
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