What is the GCF 16 24 40

1 answer:

4 0

The factors of 16 are 1, 2, 4, 8, and 16.
The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. 
The factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40. 
So the greatest common factor of 16, 24, and 40 is 8.

Hope this helps:)

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2.8y + 6 +0.2 y =5y -14

serious [3.7K]

2.8y + 6 + 0.2y = 5y - 14
(Simplify like terms in first half of equation)
3y + 6 = 5y - 14
(Subtract both sides by 3y)
6 = 2y - 14
(Flip)
2y - 14 = 6
(Add both sides by 14)
2y = 20
(Divide both sides by 2)
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3 years ago

A complex shape with measurements 10 feet, 4 feet, 7 feet, 5 feet, 3 feet, 9 feet.

Archy [21]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Find the roots of:

Elena-2011 [213]

$\large\displaystyle\text{$\begin{gathered}\sf \pmb{1) \ 2x^3-7x^2+8x-3=0 } \end{gathered}$}$

Synthetic division is used since the equation is of the third degree. The divisors of -3 are 1, -1, 3, +3. So:

| 2 -7 8 -3

1 | 2 -5 3

| 2 -5 3 0

1 | 2 -3

2 -3 0

So the factorization is (x-1)² (2x-3)=0. So:

$\bf{ x_1=x_2=1 \qquad x_2=\dfrac{3}{2} }$

$\large\displaystyle\text{$\begin{gathered}\sf \pmb{2) \ x^3-x^2-4=0 } \end{gathered}$}$

Synthetic division is used since the equation is of the third degree. The divisors of -4 are 1, -1, 2, -2, 4, -4. So:

| 1 -1 0 -4

2 | 2 2

1 2 2 0

So the factorization is (x-2)(x²+x+2)=0 . When calculating the discriminant of the trinomial, it is concluded that it has no roots since the result is negative. So you only have one solution.

$\bf{ 1^2-4(2)(2)=1-16=-15 < 0 \quad \Longrightarrow \quad x=2 }$

$\large\displaystyle\text{$\begin{gathered}\sf \pmb{3) \ 6x^3+7x^2-9x+2=0 } \end{gathered}$}$

Synthetic division is used since the equation is of the third degree. The divisors of 2 are 1, -1, 2, -2. So:

| 6 7 9 2

-2 | -12 10 -2

6 -5 1 0

So the factorization is (x+2)(6x²-5x+1)=0 . The quadratic equation is solved by the general formula:

$\bf{ x_{2, 3}&=\dfrac{5\pm \sqrt{(5)^2-4(6)(1)}}{2(6)}=\dfrac{5\pm \sqrt{25-24}}{12}=\dfrac{5\pm 1}{12} }}$

$\large\displaystyle\text{$\begin{gathered}\sf \begin{matrix} x_1=-2&\ \ \ \ \ \ x_{2}=\dfrac{6}{12} \qquad &\ \ \ x_3=\dfrac{4}{12}\\ &\ \ \ x_2=\dfrac{1}{2} \qquad &x_3=\dfrac{1}{3} \end{matrix} \end{gathered}$}$