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3 years ago

Find the least common denominator between the two fractions. 1/4 and 3/5

1 answer:

6 0

To Find the least common denominator you have to find the multiples of each number

multiple means the product of each factor

so the multiples of 4 would be 4 8 12 16 20

you want to name at least five multiples but if there are not the same multiples you might have to list more

the multiples of 5 would be 5 10 15 20

then I can stop there because I have found two of the same multiples

so now we have our least common denominator which is 20

HOPE THIS HELPS!!!

You might be interested in

Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

5 0

3 years ago

Pretest: Unit 1

Alex_Xolod [135]

A vertical line that the graph of a function approaches but never intersects. The correct option is B.

<h3>When do we get vertical asymptote for a function?</h3>

Suppose that we have the function f(x) such that it is continuous for all input values < a or > a and have got the values of f(x) going to infinity or -ve infinity (from either side of x = a) as x goes near a, and is not defined at x = a, then at that point, there can be constructed a vertical line x = a and it will be called as vertical asymptote for f(x) at x = a

A vertical asymptote can be described as a vertical line that the graph of a function approaches but never intersects.

Hence, the correct option is B.

Learn more about Vertical Asymptotes:

brainly.com/question/2513623

#SPJ1

3 0

2 years ago

Harry Joodini has insured his home for 85% of its replacement value of $185,000. Use the chart to determine the amount of covera

Reika [66]

Total coverage = 0.85 x 185,000 = $157,250

Coverage for Loss of Use = 0.25 x 157250 = $39,312.50

7 0

3 years ago