All categories

3 years ago

LCM and the GCF of 12 and 32?

Mathematics

1 answer:

OverLord2011 [107]3 years ago

7 0

The greatest common factor of 12 and 32 is 4.

The least common multiple of 12 and 32 is 96.

Hope I helped☺☺

You might be interested in

What is the 158th term of the sequence 8,6,4..?

Harman [31]

Answer:

-306

Step-by-step explanation:

... -300, -302, -304, -306, -308, -310, -312 ...

7 0

3 years ago

Francis works at Carlos Bakery and is making cookie trays. She has 48 chocolate chip cookies, 64 rainbow cookies, and 120 oatmea

amm1812

The number of cookies and trays are illustrations of greatest common factors.

The number of trays is 8
6 chocolate chips, 8 rainbows and 15 oatmeal cookies would fit each tray

The given parameters are:

$\mathbf{Chocolate\ chip=48}$

$\mathbf{Rainbow=64}$

$\mathbf{Oatmeal=120}$

(a) The number of trays

To do this, we simply calculate the greatest common factor of 48, 64 and 120

Factorize the numbers, as follows:

$\mathbf{48 = 2 \times 2 \times 2 \times 2 \times 3}$

$\mathbf{64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2}$

$\mathbf{120 = 2 \times 2 \times 2 \times 3 \times 5}$

So, the GCF is:

$\mathbf{GCF= 2 \times 2 \times 2}$

$\mathbf{GCF= 8}$

Hence, the number of tray is 8

(b) The number of each type of cookie

We have

$\mathbf{Chocolate\ chip=48}$

$\mathbf{Rainbow=64}$

$\mathbf{Oatmeal=120}$

Divide each cookie by the number of trays

So, we have:

$\mathbf{Chocolate\ chip = \frac{48}{8} = 6}$

$\mathbf{Rainbow = \frac{64}{8} = 8}$

$\mathbf{Oatmeal = \frac{150}{8} = 15}$

Hence, 6 chocolate chips, 8 rainbows and 15 oatmeal cookies would fit each tray

Read more about greatest common factors at:

brainly.com/question/11221202

4 0

2 years ago

7. Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0

morpeh [17]

Hey there!

Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0

Answer :

x = -4 or x = 2 ✅

Explanation :

Quadratic formula : ax² + bx + c = 0 where a ≠ 0

The number of real-number solutions (roots) is determined by the discriminant (b² - 4ac) :

If b² - 4ac > 0 , There are 2 real-number solutions

If b² - 4ac = 0 , There is 1 real-number solution.

If b² - 4ac < 0 , There is no real-number solution.

The roots of the equation are determined by the following calculation:

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

Here, we have :

a = 1
b = 2
c = -8

1) Calculate the discriminant :

b² - 4ac ⇔ 2² - 4(1)(-8) ⇔ 4 - (-32) ⇔ 36

b² - 4ac = 36 > 0 ; The equation admits two real-number solutions

2) Calculate the roots of the equation:

▪️ (1)

$x_1 = \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x_1 = \frac{ - 2 - \sqrt{36} }{2(1) } \\ \\ x_1 = \frac{ - 2 - 6}{2} \\ \\ x_1 = \frac{ - 8}{2} \\ \\ \blue{\boxed{\red{x_1 = -4}}}$

▪️ (2)

$x_2 = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ x_2 = \frac{ - 2 + \sqrt{36} }{2(1)} \\ \\ x_2 = \frac{ - 2 + 6}{2} \\ \\ x_2 = \frac{4}{2} \\ \\ \red{\boxed{\blue{x_2 = 2}}}$