All categories

3 years ago

Why is the LCM always greater than the GCF?

Mathematics

1 answer:

AlladinOne [14]3 years ago

5 0

Because the gcf is 12 and the lcm is 1000000000000000000000000000000000000000000000000000

You might be interested in

Arrange the equations in increasing order of the value of their solutions.

DiKsa [7]

Answer:

The answer is below

Step-by-step explanation:

1/4x + 5/2x - 2 = 4 - 1/4x

collecting like terms:

(1/4)x + (5/2)x + (1/4)x = 4 + 2

3x = 6

x = 6/3

x = 2

7.9x + x + 4 = -1.1x - 16

collecting like terms:

7.9x + x + 1.1x = - 16 - 4

10x = -20

x = -2

3.2x + 5.7 = -2.5x

collecting like terms:

3.2x + 2.5x = 5.7

5.7x = 5.7

x = 1

10.1x - 1.6x + 44 = -7

collecting like terms:

10.1x - 1.6x = -7 - 44

8.5x = -51

x = -6

Hence the equations in increasing order of the value of their solutions is:

10.1x - 1.6x + 44 = -7

7.9x + x + 4 = -1.1x - 16

3.2x + 5.7 = -2.5x

1/4x + 5/2x - 2 = 4 - 1/4x

4 0

3 years ago

Read 2 more answers

Need help with this math question please

lesya692 [45]

Answer:

See below for answer.

Step-by-step explanation:

RP/RT =RQ/RS Given

∠R = ∠R

ΔPQR similar to Δ TSR If the measures of two sides of a triangle are proportional to the measures of two corresponding sides of another triangle and their included angles are congruent, then the triangles are similar.

3 0

3 years ago

Pls help . . . . . . . . . . .

Korolek [52]

Answer:

8×8 then 2×2 put those answers together and divide more answers at photomath

Step-by-step explanation:

×8

×2

64÷4=?

8 0

3 years ago

A child wanders slowly down a circular staircase from the top of a tower. With x,y,zx,y,z in feet and the origin at the base of

babymother [125]

Answer:

a) The tower is 90 feet tall

b) She reaches the bottom at t = 18 minutes.

c) Her speed at time t is 5 \sqrt[]{5} ft/minute

d) Her acceleration at time t is 10 ft/minute^2

Step-by-step explanation:

Consider the path described by the child as going down the tower to have the following parametrization $\gamma(t) = (10\cos t, 10 \sin t, 90-5t)$

a) Assuming that the child is at the top of the tower when she starts going down, we have that at the initial time (t=0) we will have the value of the height of the tower. That is z = 90-5*0 = 90 ft.

b) The child reaches the bottom as soon as z =0. We want to find the value of t that does that. Then we have 0 = 90-5t, which gives us t = 18 minutes.

c) Given the parametrization we are given, the velocity of the child at time t is given by $\frac{d\gamma}{dt}= (\frac{d}{dt}(10\cos t), \frac{d}{dt} (10 \sin t ), \frac{d}{dt}(90-5t)) = (-10 \sin t, 10 \cos t, -5)$ . The speed is defined as the norm of the velocity vector,

so, the speed at time t is given by $v = \sqrt[]{(-10 \sin t)^2+(10 \cos t)^2+(-5)^2} = \sqrt[]{100(\sin^2 t + \cos^2 t)+25} = \sqrt[]{125}= 5 \sqrt[]{5}$