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notka56 [123]
3 years ago
12

How do I factor trinomials of a form ax^2+bx+c when a=1 In one paragraph

Mathematics
1 answer:
vitfil [10]3 years ago
6 0

Answer:

  • Trinomials in the form x^2+bx+c can often be factored as the product of two binomials.

Step-by-step explanation:

As we know that a polynomial with three terms is said to be a trinomial.

Considering the trinomial of a form

ax^2+bx+c

As

a = 1

so

x^2+bx+c

  • Trinomials in the form x^2+bx+c can often be factored as the product of two binomials.

For example,

x^2+7x+10

=\left(x^2+2x\right)+\left(5x+10\right)

=x\left(x+2\right)+5\left(x+2\right)

\mathrm{Factor\:out\:common\:term\:}x+2

=\left(x+2\right)\left(x+5\right)

Therefore, Trinomials in the form x^2+bx+c can often be factored as the product of two binomials.

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In 1993, funding for a program increased by 0.79 billion dollars from the funding in 1992. In 1994, the increase was 0.49 billio
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Hi I'm new wanna be my friend =D!!!!!!!!!!!!!
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3 years ago
The coefficients of a, b, and c in the following quadratic equation are a = 4, b = 4, and c = -1:
iVinArrow [24]

if c is -1 the equation should be

4x² + 4x -1 = 0

not plus 1

I assume you want the coordinates of the zeros.

if c is indeed -1, they are at (-1.207 , 0) and (0.207 , 0)

if c is +1, as in the equation the only zero point is at (-0.5 , 0)

see screenshot for illustration

5 0
3 years ago
Round 1.5 to one decimal place
xxMikexx [17]
If the last digit in 1.50 is less than 5, then remove the last digit.

If the last digit in 1.50 is 5 or more and the second to the last digit in 1.50 is less than 9, then remove the last digit and add 1 to the second to the last digit.

If the last digit in 1.50 is 5 or more and the second to the last digit in 1.50 is 9, then remove the last digit, make the second to last digit 0, and add 1 to the number to the left of the decimal place.

When rounding 1.50 to one decimal place we use One Decimal Place Rule #1. Therefore, the answer to "What is 1.50 rounded to 1 decimal place?" is:

1.5
3 0
3 years ago
Read 2 more answers
How many ways can we pick three distinct integers from {1, 2, ..., 15} such that their sum is divisible by 3?
gogolik [260]

Answer:

Clearly from 1 to 15 we can only have the sum 6,9,12 and 15 to be divisible by 3.

Now meaning we have 4! ( 4 factorial) which implies 4x3x2x1= 24 .

We have 24 ways to pick distinct integers that are divisible by 3.

Step-by-step explanation:

The first thing you need to do is to look for numbers that are divisible by 3. Then you take its factorial

5 0
3 years ago
At homecoming, 125 tickets where sold. Tickets purchased for homecoming cost $20.5 without an activity card, and $15.75 with an
ale4655 [162]

Answer:

Nearly 83 students did not had activity card.

Nearly 42 students had an activity card.

Step-by-step explanation:

Total number of tickets sold at homecoming = 125

Cost of each  ticket without activity card = $20.5

Cost of each  ticket with activity card = $15.75

Total amount earned  = $2365

Let, the number of tickets sold without activity car d  = k

So, the number of tickets sold without card = (Total sold - k)

                                                                          = 125 - k

Now, according to the question:

$20.5(k) + $15.75(125 - k)  =  $2365

or, 20.5 k + 1968.5 - 15.75 k = 2365

⇒ 4.75 k = 2365 - 1968.5 = 396.5

⇒ k = \frac{396.5}{4.75 }  = 83.47

Hence, nearly 83 students did not had activity card.

So,the number of students that had activity card = 125 - 83.47 = 41.53 ~  42

Hence, nearly 42 students had an activity card.

4 0
3 years ago
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