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babymother [125]
3 years ago
7

What are the factors of 3x2 + 10x + 3? (3x − 1)(x − 3) (3x − 3)(x − 1) (3x + 1)(x + 3) (3x + 3)(x + 1)

Mathematics
2 answers:
natka813 [3]3 years ago
4 0
The factors of 3x^2+10x+3 are (3x+1)(x+3)
finlep [7]3 years ago
3 0

Answer:

(3x+1)(x+3)

Step-by-step explanation:

We have been given an expression 3x^2+10x+3. We are asked to factor the given expression.

We will factor our given expression by splitting the middle term. We need two numbers which add up-to 10 and whose product is 9.

We know that 1 and 9 are such numbers.

3x^2+9x+1x+3

Make two groups:

(3x^2+9x)+(1x+3)

3x(x+3)+1(x+3)

(3x+1)(x+3)

Therefore, our required factors are (3x+1)(x+3) .

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  First, you try to make sure that both the denominators are the same. In order to do this, you multiply both fractions' numerator and denominator by the other fraction's denominator. Next, you combine like terms in the numerator, and then you simplify the fraction.
8 0
3 years ago
Not really sure how to do this with a negative number
Oksi-84 [34.3K]
1/2*(-1/7) =  - (1*1)/(2*7)= - 1/14

(+)*(-) = (-)

Answer : - 1/14
4 0
3 years ago
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piper needs to buy 320 new ping-pong balls for the team ping pong balls come in 16 how many sets of ping-pong balls should piper
Lelu [443]

Answer:

20 sets

Step-by-step explanation:

320/16=20

He needs 320 ping pong balls, and he has 16, so divide the amount needed by the amount that comes in each containter.

3 0
3 years ago
What is the circumference of a circle with a diameter of 2.9 cm?​
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7 0
3 years ago
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A 2-kg object is fired from a cannon at a 30º angle wrt to the horizontal. the initial speed is 35 m/s. how high will the object
kompoz [17]
     <span> You have :
----------------

v0 = 35 m/s

B = angle of elevation = 30 deg

a suby = - g

v suby0 = ( v0 ) ( sin B )

v suby0 = ( 35 m/s ) ( sin 30 ) = 17.5 m/s

At maximum height :
------------------------------

v suby = vy* = 0.0

t = t*

y = ymax = y*

Basic kinematics gives :
-------------------------------------

a suby = dv suby / dt = -g ------> dt = dv suby /a suby

v suby = dy/dt ------> dt = dy / vsuby

dt = dv suby / a suby = dy / v suby

( v suby ) ( d v suby) = ( a suby ) dy = ( -g ) dy

Now integrate and get :
-----------------------------------

( v suby* )^2 - ( v suby0)^2 ( 1/2 ) = ( -g ) ( ymax )

ymax = [ ( v suby0 )^2 - ( v suby* )^2 ] / [ ( 2 ) ( g ) ]

ymax = ( 17.5 m/s )^2 - ( 0.0 m/s )^2 ] / [ ( 2 ) ( 9.807 m/s^2 ) ]

ymax = 15.6 m <------------------

Check the result:
--------------------------

t* = v suby0 / g = ( 17.5 m/s ) / ( 9.807 m/s^2 ) = 1.784 s

y* - y0 = ( v suby0 ) ( t* ) - ( g/2 ) ( t* )^2

y* = ( 17.5 m/s ) ( 1.784 s ) - ( 9.807 m/s^2 / 2 ) ( 1.784 s )^2

y* = ymax = 31.23 m - 15.61 m = 15.6 m

The ymax values agree.</span>
6 0
3 years ago
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