Please please please help fast

using the numbers 3, 5, and 8, can you write nine proper fractions and nine improper fractions? You may use each number only onc

Bond [772]

Answer:

First, we can write a fraction as a/b

Where a is the numerator, and b is the denominator.

A proper fraction is a fraction where the numerator is smaller than the denominator.

Using only the given numbers (only once per fraction), some examples of proper fractions are:

3/5

3/8

5/8

3/85

3/58

5/83

5/38

8/35

8/53

You can see that in all of them the denominator is larger than the numerator.

The improper fractions are those where the numerator is equal or larger than the denominator.

The 9 examples using the given numbers are:

5/3

8/5

8/3

35/8

38/5

53/8

58/3

83/5

85/3

6 0

2 years ago

Please solve all with explanation

garri49 [273]

Answer:

1. $\frac{x^2}{y^8}$

2 a) $\frac{-49}{2}$

b) 0

3. x = 10

Step-by-step explanation:

1.

$(\frac{x^{-3}y^4}{x^{-2}})^{-2}\\({x^{-3+2}y^4})^{-2}\\({x^{-1}y^4})^{-2}({x^{-1*-2}y^{4*-2})= \frac{x^2}{y^8}$

2. a)

$2 [(\frac{-1}{4}) -12 (-1)^2] - 5[24 +2 (-12)]^2\\2[(\frac{-1}{4}) -12] - 5[24 -24]^2\\2 * [(\frac{-1}{4}) -12] -0\\\frac{-49}{2}$

b)

$-a^2b^2 - 3a^3b^2\\-\frac{1}{9}*1 - 3 * \frac{-1}{27} *1\\-\frac{1}{9} + \frac{1}{9}\\0$

3.

$\frac{3}{4}x + \frac{5}{6} = 5x -\frac{125}{3} \\\frac{5}{6} + \frac{125}{3} = 5x - \frac{3}{4}x\\\frac{5+250}{6} = \frac{20x -3x}{4}\\\frac{255}{6} = \frac{17x}{4}\\x = 10$

7 0

2 years ago

6. use the distributive property to simplify. -3(x - 10) + x

Roman55 [17]

Answer:

for 6: -2x+30

I dont know for # 8

Step-by-step explanation:

to distribute, you have to multiply -3 by x and -10.

-3(x) - -3(10) + x =

-3x + 30 + x:

then you have to add common terms

-3x + x = -2x

then replace that in your equation to get:

-2x +30

4 0

2 years ago

The position vector of p (relative to the origin) is op = (x, y). if the magnitude of op is 5 units, find the set of all possibl

Firdavs [7]

Answer:

S = {(0, 5), (0, - 5), (3, 4), (3, -4), (-3, 4), (-3, -4), (4, 3), (4, -3), (-4, 3), (-4, -3), (5,0), (-5, 0) }

Step-by-step explanation:

Remember that the distance between two points (a, b) and (c, d) is given by:

$distance = \sqrt{(a - c)^2 + (b - d)^2}$

So, here we have that the distance between the point (x, y) and the origin, (0, 0) must have a magnitude of 5 units, then we want to solve:

$5 = \sqrt{(x - 0)^2 + (y - 0)^2} \\5 = \sqrt{x^2 + y^2}$

If we square both sides, we get

$5^2 = (\sqrt{x^2 + y^2} )^2\\25 = x^2 + y^2$

Now we want to find all the points (x, y) that meet this condition, suc that:

x ∈ Z

y ∈ Z

So both x and y must be integers.

So here we can just play with different values of x and y.

For example, if we define:

x = 0 we get:

25 = 0^2 + y^2

25 = y^2

√25 = y

Then we can have y = 5 or y = -5

from this we got two points:

(0, 5) and (0, - 5)

if x = 1 we have:

25 = 1^2 + y^2

25 - 1 = y^2

24 = y^2

There is no integer such that its square is equal to 24, so we can stop here.

if x = 2 or - 2, we have:

25 = 2^2 + y^2

25 = 4 + y^2

25 -4 = 21 = y^2

Again, there is no integer such that its square is equal to 21, so we can stop here.

if x = +3 or -3, we have:

25 = 3^2 + y^2

25 = 9 + y^2

25 - 9 = 16 = y^2

√16 = y

then we can have y = 4 or y = -4

from this we got four points:

(3, 4)

(-3, 4)

(3, -4)

(-3, -4)

And for symmetry, if x = 4 or -4 we have the points:

(-4, 3)

(4, 3)

(-4, -3)

(4, -3)

finally, again for symmetry, if we take x = 5 or x = -5 we have the points:

(5,0)

(-5, 0)

Concluding, the set of all possible values (x, y) is:

S = {(0, 5), (0, - 5), (3, 4), (3, -4), (-3, 4), (-3, -4), (4, 3), (4, -3), (-4, 3), (-4, -3), (5,0), (-5, 0) }

7 0

3 years ago

Rectangle QRST has vertices Q(3,2) and S(7,8) What are two possible coordinates for vertices R and T

Bumek [7]

Check the picture below.

6 0

2 years ago