All categories

4 years ago

Find the LCM (least common multiple) of 25 and 35. LCM =

Mathematics

1 answer:

Paladinen [302]4 years ago

5 0

Answer:

175

Step-by-step explanation:

Hi there !

25 = 5²

35 = 5×7

Lcm = multiply together each of the prime numbers with the highest power

Lcm(25 ; 35) = 5²×7 = 25×7 = 175

Good luck !

You might be interested in

Please help help ASAP ASAP

Elenna [48]

Answer:

Step-by-step explanation:

6 0

3 years ago

Please help me with 2 & 4

Serjik [45]

2) 1320cm squared because you would multiply length x width x height

4) 11.132ft squared because again you multiply length x width x height

6 0

3 years ago

Read 2 more answers

Is brainly lacking features?

AlekseyPX

brainly should let more people answer the question's

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

The cost, C, of producing x-bikes is given by C = 22x +132 . The revenue gained from selling x-bikes is given by R=350x. Ifthepr

Butoxors [25]

Answer:

I think its B

Step-by-step explanation:

5 0

3 years ago

A quadratic equation is shown below: 3x2 − 15x + 20 = 0 Part A: Describe the solution(s) to the equation by just determining the

Triss [41]

Answer:

$\text{The roots of }3x^2+5x-8=0\text{ are }x=1,\frac{-8}{3}$

Step-by-step explanation:

$\text{Part A: Given a quadratic equation }3x^2-15x+20=0$

$\text{Comparing above equation with }ax^2+bx+c=0$

a=3, b=-15, c=20

Discriminant can be calculated as

$D=b^2-4ac$

$D=(-15)^2-4(3)(20)=225-240=-15$

The roots are imaginary

The solution is

$x=\frac{-b\pm\sqrt{D}}{2a}$

$x=\frac{-(-15)\pm \sqrt{-15}}{2(3)}=\frac{15\pm\sqrt{15}i}{6}$

The roots are not real i.e these are imaginary

$\text{Part B: Given a quadratic equation }3x^2+5x-8=0$

$\text{Comparing above equation with }ax^2+bx+c=0$

a=3, b=5, c=-8

Discriminant can be calculated as

$D=b^2-4ac$

$D=(5)^2-4(3)(-8)=25+96=121>0$

The roots are real

By quadratic formula method

The solution is

$x=\frac{-b\pm\sqrt{D}}{2a}$

$x=\frac{-5)\pm \sqrt{121}}{2(3)}=\frac{-5\pm 11}{6}$

$x=1,\frac{-8}{3}$

which are required roots.

I choose this method because I can get the solutions directly by substituting the values in formula, and I don't have to guess the possible solutions.

6 0

3 years ago