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mr_godi [17]
3 years ago
8

Whats p divided by 3

Mathematics
1 answer:
Arada [10]3 years ago
4 0
P over 3 is the answer
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1) Given the equation ax + 3 2 = 6, solve for x.
ruslelena [56]
Ax + 32 =  6
     - 32  = -32
---------------------
 ax         = -26
----            -----
 a                a

1x          =  -26
                  -----
                     a

x             = -26
                  -----
                     a

4 0
2 years ago
Read 2 more answers
Each of the following equations is an example of one of the properties of whole-number addition. Fill in the black to make a tru
Oliga [24]

a. 4 – commutative property

b. 5 – commutative property

c. 0 – identity property

d. 4 – associative property

_____

The commutative property lets you swap the order: (a) + (b) = (b) + (a).

The associative property lets you change the grouping: (a+b)+c = a+(b+c).

The identity property lets you add 0 without changing anything: (a) +0 = (a).

5 0
2 years ago
I need help with this
snow_lady [41]

Answer:

x = 2 square roots of 2

Step-by-step explanation:

it is a 45-45-90 triangle and in that the two legs are equal to each other so both are 2 and then the hypotenuse is x times the square root of 2

8 0
2 years ago
40 POINTS!!! WILL MAKE BRAINLIEST FOR FIRST ANSWER TO BE CORRECT!
vredina [299]

Answer:

3

Step-by-step explanation:

We want to basically find all the points at which the line y = 3 intersects the given graph of g(x).

y = 3 is simply a straight horizontal line intersecting the y-axis at (0, 3). If we draw this line, we will see that it intersects g(x) at 3 different points. This indicates that there are 3 values of x where g(x) = 3.

Hope this helps!

6 0
3 years ago
Read 2 more answers
Find the derivative of f(x)=x+1/x-1 using first principal​
zepelin [54]

f ' ( x ) = 1 ( x + 1 ) 2

 

Explanation:

differentiating from first principles

f ' ( x ) = lim h → 0

 

f ( x + h ) − f ( x ) h

f ' ( x ) = lim h → 0

 x + h x + h + 1 − x x + 1 h

the aim now is to eliminate h from the denominator

f ' ( x ) = lim h =0  

( x + h ) ( x + 1 )− x ( x + h + 1) h ( x + 1 ) ( x + h + 1 )

f ' ( x ) = lim h → 0

 x 2 + h x + x + h − x 2 − h x − x h ( x + 1 ) ( x+h + 1 )

f ' ( x ) = lim h → 0

 

h 1 h 1 ( x + 1 ) ( x + h +1 )

f ' ( x ) = 1 ( x + 1 ) 2

5 0
3 years ago
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