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scZoUnD [109]
3 years ago
6

What are factors of 13

Mathematics
2 answers:
andrezito [222]3 years ago
7 0
Answer: 1, 13 is the answer
RoseWind [281]3 years ago
3 0

1 and 13 are the only factors of 13.

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3
Svet_ta [14]

Answer:17,000

Step-by-step explanation:Add 532+150=680 then times it by 25 which equals 17,000

5 0
3 years ago
Read 2 more answers
Solve the system of equations. -5x+6y=40;3y-8x=-46
Effectus [21]

Equation Form:

x

=

12

,

y

=

50

3

or

x

=

12

,

y

=

50

3

x

=

12

,

y

=

50

3

x

=

12

,

y

=

50

3

x

=

12

,

y

=

50

3

Step-by-step explanation: I have no idea why it looks like that

8 0
3 years ago
When factoring a polynomial in the form ax2 + bx + c, where a, b, and c are positive real numbers, should the signs in the binom
gtnhenbr [62]

When factoring a polynomial in the form ax2 + bx + c, where a, b, and c are positive real numbers, the signs in the binomials should be both positive

<h3>What are quadratic equations?</h3>

Quadratic equations are second-order polynomial equations and they have the form y = ax^2 + bx + c or y = a(x - h)^2 + k

<h3>How to determine the true statement?</h3>

The form of the polynomial is given as:

ax2 + bx + c

Where a, b, and c are positive real numbers.

Since a, b, and c are positive real numbers. then the form of the expansion would be:

ax2 + bx + c = (dx + e)(fx + g)

<h3>Example to verify the claim</h3>

Take for instance, we have the following quadratic equation

x^2 + 6x +  8

Expand the equation

x^2 + 6x +  8 = x^2 + 4x + 2x +  8

Factorize the equation

x^2 + 6x +  8 = (x + 2)(x + 4)


Hence, the signs in the binomials should be both positive

Read more about polynomials at:

brainly.com/question/4142886

#SPJ1

5 0
1 year ago
Solve -14 +5x =-12 really need this answer
Paladinen [302]
Move the 14 over and get
5x = 2
divide by 5
x = 2/5
7 0
3 years ago
Read 2 more answers
Write in simplest form 6 times 2/3
satela [25.4K]
4 
I HAVE TO WRITE STUFF OR ELSE IT WON;T LET ME POST YOUR ANSWER IS 4

8 0
3 years ago
Read 2 more answers
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