All categories

3 years ago

The product of two mixed numbers x and y is 10. If x= _ 2/3 and y=_ 3/4 , find x and y?

Mathematics

1 answer:

Akimi4 [234]3 years ago

6 0

Answer:

x = 2 2/3
y = 3 3/4

Step-by-step explanation:

There may be other ways to do this. Here's the approach I took. Let the integer part of x be represented by "a", and the integer part of y be represented by "b". Now, we have ...

(a +2/3)(b +3/4) = 10

Multiplying by 12, this becomes ...

(3a +2)(4b +3) = 120

There are only 16 divisors of 120, so they can be easily enough checked one at a time.

120 = 1·120 = 2·60 = 3·40 = 4·30 = 5·24 = 6·20 = 8·15 = 10·12

Of these divisors, the ones that are of the form 3a+2 are ...

2, 5, 8, 20 . . . . a=0, 1, 2, 6

The ones that are of the form 4b+3 are ...

3, 15 . . . . b = 0, 3

The factor pair that consists of a number on the first list and a number on the second list is ...

120 = 8·15

corresponding to a=2 and b=3. So, the mixed numbers you want are ...

x = 2 2/3

y = 3 3/4

_____

Check

(2 2/3)·(3 3/4) =2·3 +2·3/4 +2/3·3 +2/3·3/4 = 6 +6/4 +2 +2/4 = 8+8/4 = 10

You might be interested in

Marty and Ethan both wrote a function, but in different ways.

marishachu [46]

Question:

Marty and Ethan both wrote a function, but in different ways.

Marty

y+3=1/3(x+9)

Ethan

x y

-4 9.2

-2 9.6

0 10

2 10.4

Whose function has the larger slope?

1. Marty’s with a slope of 2/3

2. Ethan’s with a slope of 2/5

3. Marty’s with a slope of 1/3

4. Ethan’s with a slope of 1/5

Answer:

Marty’s with a slope of 1/3 has the larger slope

Solution:

Given that Marty equation is:

$y + 3 = \frac{1}{3}(x+9)$

The point slope form is given as:

$y - y_1 = m(x-x_1)$

Where, "m" is the slope of line

On comapring both equations,

$m = \frac{1}{3}$

Ethan wrote a function:

Consider any two values from the table we have;

(0, 10) and (2, 10.4)

The slope is given by formula:

$m = \frac{y_2-y_1}{x_2-x_1}$

From above two points,

$(x_1, y_1) = (0, 10)\\\\(x_2, y_2) = (2, 10.4)$

Therefore,

$m = \frac{10.4-10}{2-0}\\\\m = \frac{0.4}{2} \\\\m = 0.2$

Thus we get,

$\text{Slope of Ethan} < \text{Slope of Marty}$

Therefore, Marty’s with a slope of 1/3 has the larger slope

4 0

3 years ago

Read 2 more answers

17.

eduard

Answer: C. c = 23d + 11

I'm gonna go with c = 23d + 11

Step-by-step explanation: c = 23d + 11

kikdddd

17 answers

2K people helped

its c hope that helps text me back if you want to know why it is.

...................................................................................................................................................

c is your answer

hope this helps

4 0

3 years ago

Find the range for the given domain: {-2, -1, and 0} and the function is y= 3x^2

likoan [24]

Answer:

{0, 3, 12}

Step-by-step explanation:

Put the domain values in the function and evaluate. The result is the range values.

y = 3{-2, -1, 0}^2 = 3{4, 1, 0} = {12, 3, 0}

For the given domain, the range is {0, 3, 12}.

_____

Additional comment

Neither the domain nor range values need to be put in any particular order. However, it is often convenient for them to be arranged from least to greatest.

6 0

3 years ago

Write an equation for a line on the graph that passes through the points (0.4) and (12,16)

Montano1993 [528]

Answer:

$y = x + 4$

Step-by-step explanation:

Use the two-point form of the equation of a line.

$y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

$y - 4 = \dfrac{16 - 4}{12 - 0}(x - 0)$

$y - 4 = \dfrac{12}{12}x$

$y - 4 = x$

$y = x + 4$

3 0

3 years ago

Read 2 more answers

[20points][Precal] Find the cross product of -(3/4)V and (1/2)w if v = <-2, 12, -3> and w = <-7, 4, -6>

Lana71 [14]

$\mathbf v=\langle-2,12,-3\rangle=-2\,\vec i+12\,\vec j-3\,\vec k$
$-\dfrac34\mathbf v=\dfrac32\,\vec i-9\,\vec j+\dfrac94\,\vec k$

$\mathbf w=\langle-7,4,-6\rangle=-7\,\vec i+4\,\vec j-6\,\vec k$
$\dfrac12\mathbf w=-\dfrac72\,\vec i+2\,\vec j-3\,\vec k$

$-\dfrac34\mathbf v\times\dfrac12\mathbf w=\begin{vmatrix}\vec i&\vec j&\vec k\\\frac32&-9&\frac94\\-\frac72&2&-3\end{vmatrix}$
$=\left((-9)\times(-3)-\dfrac94\times2\right)\,\vec i-\left(\dfrac32\times(-3)-\dfrac94\times\left(-\dfrac72\right)\right)\,\vec j+\left(\dfrac32\times2-(-9)\times\left(-\dfrac72\right)\right)\,\vec k$
$=\dfrac{45}2\,\vec i-\dfrac{27}8\,\vec j-\dfrac{57}2\,\vec k=\left\langle\dfrac{45}2,-\dfrac{27}8,-\dfrac{57}2\right\rangle$

6 0

3 years ago