All categories

3 years ago

X÷0.4=2 1/9 ÷3 1/6 solve for x

Mathematics

1 answer:

Diano4ka-milaya [45]3 years ago

5 0

Answer:

x = 0.27

Step-by-step explanation:

Given:

x÷0.4=2 1/9 ÷3 1/6

To Find:

x = ?

Solution:

First we will see what is

$2\frac{1}{9}= \frac{2\times 9 + 1}{9} \\\\2\frac{1}{9}= \frac{18 + 1}{9} \\\\2\frac{1}{9}= \frac{19}{9} \\\\Similarly,\\\\3\frac{1}{6}= \frac{3\times 6 + 1}{6} \\\\3\frac{1}{6}= \frac{18 + 1}{6} \\\\3\frac{1}{6}= \frac{19}{6} \\$

Now the final equation is

$\frac{x}{0.4} =\frac{\frac{19}{9}}{\frac{19}{6}}\\\\\frac{x}{0.4} =\frac{6}{9}\\\\\frac{x}{0.4} =\frac{2}{3}\\\\\therefore x = 0.4\times \frac{2}{3}\\\\x = \frac{0.8}{3}\\\\x=0.2666\\\\x=0.27$

x = 0.27

You might be interested in

HELP!!!!!!!!!!!!!!!!!!!!

Sholpan [36]

Answer:

pretty easy two

Step-by-step explanation:

ye just look

4 0

3 years ago

Read 2 more answers

Prove the proposition p (0), where p (n) is the proposition "if n is a positive integer greater than 1, then n2 > n." what ki

expeople1 [14]

Answer:

Vacuous proof is used.

Step-by-step explanation:

Given:

Proposition p(n) :

"if n is a positive integer greater than 1, then n² > n"

To prove:

Prove the proposition p (0)

Solution:

Using the proposition p(n) the proposition p(0) becomes:

p(0) = "if 0 is a positive integer greater than 1, then 0² > 0"

The proposition that "0 is a positive integer greater than 1" is false

Since the premises "if 0 is a positive integer greater than 1" is false this means the overall proposition/ statement is true.

Thus this is the vacuous proof which states that:

if a premise p ("0 is a positive integer greater than 1") is false then the implication or conditional statement p->q ("if n is a positive integer greater than 1, then n² > n") is trivially true.

So in vacuous proof, the implication i.e."if n is a positive integer greater than 1, then n2 > n." is only true when the antecedent i.e. "0 is a positive integer greater than 1" cannot be satisfied.

6 0

3 years ago

A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4

Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$