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3 years ago

Find the HIGHEST Common and lowest common multple of 36 and 48 using prime factors

Mathematics

2 answers:

Fed [463]3 years ago

7 0

Answer:

the least is 144 the highest is 12

Step-by-step explanation:

Vesnalui [34]3 years ago

4 0

Answer:

Step-by-step explanation:

Least common multiple can be found by multiplying the highest exponent prime factors of 36 and 48. First we will calculate the prime factors of 36 and 48.

Prime Factorization of 36

Prime factors of 36 are 2, 3. Prime factorization of 36 in exponential form is:

36 = 22 × 32

Prime Factorization of 48

Prime factors of 48 are 2, 3. Prime factorization of 48 in exponential form is:

48 = 24 × 31

Now multiplying the highest exponent prime factors to calculate the LCM of 36 and 48.

LCM(36,48) = 24 × 32

LCM(36,48) = 144

List of positive integer factors of 36 that divides 36 without a remainder.

1, 2, 3, 4, 6, 9, 12, 18

Factors of 48

List of positive integer factors of 48 that divides 36 without a remainder.

1, 2, 3, 4, 6, 8, 12, 16, 24

Greatest Common Factor Number

We found the factors and prime factorization of 36 and 48. The biggest common factor number is the GCF number.

So the greatest common factor 36 and 48 is 12.

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3 years ago

What's the recursive formula of the geometric sequence below?

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3 years ago

Determine the equation of the line that passes through the points of intersection of the graphs of the quadratic functions f(x)

timofeeve [1]

Answer: $x-2y-2=0$

Step-by-step explanation:

Given :

$f(x) = x^2 - 4 \\ g(x) = - 3x^2 + 2x + 8$

Point of intersection :

$f(x)=g(x)\\x^2-4=-3x^2+2x+8\\4x^2-2x-12=0\\2x^2-x-6=0\\2x^2-4x+3x-6=0\\2x(x-2)+3(x-2)=0\\(x-2)(2x+3)=0\\x=2\,,\,\frac{-3}{2}$

$x=2\,;f(2)=2^2-4=0\\x=\frac{-3}{2}\,; f\left ( \frac{3}{2} \right )=\left ( \frac{3}{2} \right )^2-4=\frac{-7}{4}=-1.75$

So, we have points $\left ( 2,0 \right )\,,\,\left ( -1.5,-1.75\ \right )$

Equation of line passing through two points $\left ( x_1,y_1 \right )\,,\,\left ( x_2,y_2 \right )$ is given by $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left ( x-x_1 \right )$

Let $\left ( x_1,y_1 \right )=\left ( 2,0 \right )\,,\,\left ( x_2,y_2 \right )=\left ( -1.5,-1.75\ \right )$

So, equation is as follows :

$y-0=\frac{-1.75-0}{-1.5-2}\left ( x-2 \right )\\y=\frac{-1.75}{-3.5}\left ( x-2 \right )\\y=\frac{1}{2}(x-2)\\2y=x-2\\x-2y-2=0$

4 0

4 years ago

The options for number two is

Blizzard [7]

The area of any quad = the base* height
for q 1:
the base=12 in
the height= 3 in
then the area = 12*3 = 36 sq in

for q 2:
the base= 14 in
the height = 4.5 in
then the area =4.5*14 =63

3 0

3 years ago