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3 years ago

What is 0.4 as a fraction in lowest term

Mathematics

2 answers:

Colt1911 [192]3 years ago

6 0

4/10 because it isnt a 1 so it will be less

miss Akunina [59]3 years ago

3 0

2/5 is the correct answer

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Do you know the inequality? Please explain and show answer :)

DerKrebs [107]

If u have a open circle, then the inequality has no equal sign. But if it is a closed circle, the inequality has an equal sign.
Shading to the left means " less then " .
Shading to the right means " greater then " .

(1) u have an open circle on -7 with shading to the left....
x < -7

(2) u have a closed circle on 4.5 with shading to the left....
x < = 4.5 (thats less then or equal)

(3) u have an open circle on -5 with shading to the right...
x > -5

(4) u have an open circle on 1.5 with shading to the right...
x > 1.5

4 0

3 years ago

PLEASE HELP!!!!<br><br> Question below.

aleksley [76]

Step-by-step explanation:

Answer is in the pic above

5 0

2 years ago

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago

the relationship between money earned and hours worked is linear. joe computes the slope between (4, 30) and (12, 90), then comp

julia-pushkina [17]

Slope = rise/run =[y2-y1]/[x2-x1]

1) slope = [90 - 30] / [12 - 4] = 60 / 8 = 15/2

2) slope = [75 - 30] / [10 - 4] = 45 / 6 = 15 / 2

The slopps are equal.

6 0

3 years ago

Read 2 more answers

(-x+4/4)+(-x+2/4) what’s the answer?

ss7ja [257]

Answer: -2x + 3/2

Step-by-step explanation:

Simplify the expression.

3 0

3 years ago

Read 2 more answers