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frez [133]
3 years ago
6

(3x^2+2x-3)(x-1) please help

Mathematics
1 answer:
cupoosta [38]3 years ago
5 0

Answer:

Step-by-step explanation:

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

Expanding the brackets

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

3x^3 +2x^2 -3x -3x^2 -2x + 3

3x^3 - x^2 - 5x + 3

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How do you factor y=6x^2-x-2
kolbaska11 [484]

6 x -2 = -12

-12 = 3, -4

y = 6x^2 + 3x - 4x - 2

y = 3x(2x + 1) - 2(2x + 1)

(2x + 1) (3x - 2) = 0

2x = -1, 3x = 2

x = -1/2, 2/3

pls give brainliest im trying to level up :))

3 0
3 years ago
Read 2 more answers
A. 1/3=0.33<br>B. 1/6=0.17<br>C. 1/10=0.10<br>D. 1/4=0.25​
statuscvo [17]
B

lmk if this helped
4 0
3 years ago
An expression involves subtracting two numbers from a positive number. Under what circumstances with the value expression be neg
Svetach [21]
So (positive integer)-x-y can be negative
so to get a negative number as your result/answer, x+y must be greater than (positve integer) so  for example


lets say (positive integer)=8
x+y must be greater than 8 to make the equation negative so
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x=5
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8-9=-1
7 0
3 years ago
A freezer can be bought on hire purchase by making a desposiy of 15% of the cash price which is $2975 the interest which is 20%
Liono4ka [1.6K]

Answer: See explanation

Step-by-step explanation:

a. The deposit

= 15% × $2975

= 15/100 × $2975

= 0.15 × $2975

= $446.25

b. The hire purchase price

Outstanding balance = $2975 - $446.25 = $2528.75

20% of outstanding balance

= 20% × $2528.75

= 0.2 × $2528.75

= $505.75

Hire purchase price

= $2975 + $505.75

= $3480.75

c. The difference the hire purchase price and cash price

= $3480.75 - $2975

= $505.75

d. The difference as a percentage of the cash price.

= ($505.75 / $2975) × 100

= 0.17 × 100

= 17%

5 0
3 years ago
Dwayne drove 18 miles to the airport to pick up his father and then returned home. On the return trip he was able to drive an av
Nesterboy [21]

Answer:

30 mph

Step-by-step explanation:

Let d = distance (in miles)

Let t = time (in hours)

Let v = average speed driving <u>to</u> the airport (in mph)

⇒ v + 15 = average speed driving <u>from</u> the airport (in mph)

Using:  distance = speed x time

\implies t=\dfrac{d}{v}

Create two equations for the journey to and from the airport, given that the distance one way is 18 miles:

\implies t=\dfrac{18}{v}  \ \ \textsf{and} \ \  t=\dfrac{18}{v+15}

We are told that the total driving time is 1 hour, so the sum of these expressions equals 1 hour:

\implies \dfrac{18}{v} +\dfrac{18}{v+15}=1

Now all we have to do is solve the equation for v:

\implies \dfrac{18(v+15)}{v(v+15)} +\dfrac{18v}{v(v+15)}=1

\implies \dfrac{18(v+15)+18v}{v(v+15)}=1

\implies 18(v+15)+18v=v(v+15)

\implies 18v+270+18v=v^2+15v

\implies v^2-21v-270=0

\implies (v-30)(v+9)=0

\implies v=30, v=-9

As v is positive, v = 30 only

So the average speed driving to the airport was 30 mph

(and the average speed driving from the airport was 45 mph)

8 0
2 years ago
Read 2 more answers
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