Ask question

Ask question

All categories

3 years ago

9

Which is greater 1/3 or 3/8

2 answers:

morpeh [17]3 years ago

7 0

1/3 and 3/8
First, you would need to find the lowest common multiple.
It would be 24.
Change both fractions to fractions with a denominator of 24
1/3*8=8/24
3/8*3=9/24
You can see that 9 is greater that 8. Therefore 3/8 is greater.

vredina [299]3 years ago

5 0

3/8 is greater becuase when you make a common denominator of 24, 1/3 would be 8/24 and 3/8 would be 9/24. So 9/24 is greater than 8/24.

You might be interested in

An arithmetic sequence is defined by the recursive formula t1 = 11, tn = tn - 1 - 13, where n ∈N and n > 1. Which of these is

galben [10]

coolio

$t_1=11$

$t_n=t_{n-1}-13$

so each term is ound by subtracting 13 from the previous term

an aritmetic sequence can be written as

$t_n=t_1+d(n-1)$ were

$t_n$ is the nth term

$t_1$ is the first term

d is common difference, which can also be found by doing $t_n-t_{n-1}=d$

n=wich term

we know that $t_1=11$ and we can find d

$t_n=t_{n-1}-13$ , $t_n-t_{n-1}=-13=d$

so te general term is $t_n=11-13(n-1)$ which can also be expanded and written as $t_n=-13n+24$

6 0

4 years ago

What is the result if 3x-9 is evaluated when x=-5

Reptile [31]

3(-5)-9 = -24 plug in -5 for x

4 0

3 years ago

Read 2 more answers

PLZ HELP ASAP!! Drag and drop the constant of proportionality into the box to match the table. If the table is not proportional,

Morgarella [4.7K]

Answer:

not proportional

2 / 1 ≠ 3 / 2

5 0

3 years ago

This is for my friend help

Vikki [24]

Answer: 225

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4

dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

$x_1\ =\ -2$

$y_1\ =\ 3$

$z_1\ =\ -4$

Distance is measure across the line

$\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}$

So, we can write

$\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k$

$=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k$

$=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k$

$=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}$

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

$3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3$

$=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18$

$=>18k-24+12k+9+10k-32\ =\ 18$

$=>\ k\ =\dfrac{13}{8}$

So,

$x\ =\ 3k-4$

$=\ 3\times \dfrac{13}{8}-4$

$=\ \dfrac{7}{4}$

$y\ =\ \dfrac{4k+3}{2}$

$=\ \dfrac{4\times \dfrac{13}{8}+3}{2}$

$=\ \dfrac{19}{4}$

$z\ =\ \dfrac{5k-16}{3}$

$=\ \dfrac{5\times \dfrac{13}{8}-16}{3}$

$=\ \dfrac{-21}{8}$

Now, the distance of point from the plane is given by,

$d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}$

$=\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}$

$=\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}$

$=\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}$

$=\ \sqrt{\dfrac{1177}{64}}$

$=\ 4.28$

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

= 8.56 unit

So, the distance of a point from it's image is 8.56 units.

4 0

3 years ago