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2 years ago

Write 0.424242… as a fraction in simplest form

Mathematics

1 answer:

Maurinko [17]2 years ago

6 0

Answer:

14/33

Step-by-step explanation:

To write 0.424242 as a fraction you have to write 0.424242 as numerator and put 1 as the denominator. Now you multiply numerator and denominator by 10 as long as you get in numerator the whole number.

0.424242 = 0.424242/1 = 4.24242/10 = 42.4242/100 = 424.242/1000 = 4242.42/10000 = 42424.2/100000 = 424242/1000000

".424242" was replaced by "(424242/1000000)".

simplified = 212121 /500000

then i went online and got the 212121 /500000 simplified and it gave me 14/33

i hope this helps :)

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Write the equation of the line that passes through the points (- 5, 1) and (2, 0) . Put your answer in fully reduced slope inter

Doss [256]

Answer:

y=-1/7x + 12/7

Step-by-step explanation:

Start by finding the slope

m=(1-0)/(-5-2)

m=-1/7

next plug the slope and the point (-5,1) into point slope formula

y-y1=m(x-x1)

y1=1

x1= -5

m=-1/7

y- 1 = -1/7(x - -5)

y-1=-1/7(x+5)

Distribute -1/7 first

y- 1=-1/7x + 5/7

Add 1 on both sides, but since its a fraction add 7/7

y=-1/7x + (5/7+7/7)

y=-1/7x+12/7

3 0

2 years ago

(7a^3 -2a-2)+(-5a+3)

atroni [7]

Answer:

7a 3 −7a+1

Step-by-step explanation: I hope this help.

STEP

Equation at the end of step 1

((7a3 - 2a) - 2) + (3 - 5a)

STEP

Polynomial Roots Calculator :

2.1 Find roots (zeroes) of : F(a) = 7a3-7a+1

Polynomial Roots Calculator is a set of methods aimed at finding values of a for which F(a)=0

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers a which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient

In this case, the Leading Coefficient is 7 and the Trailing Constant is 1.

The factor(s) are:

of the Leading Coefficient : 1,7

of the Trailing Constant : 1

Let us test ....

P Q P/Q F(P/Q) Divisor

-1 1 -1.00 1.00

-1 7 -0.14 1.98

1 1 1.00 1.00

1 7 0.14 0.02

Polynomial Roots Calculator found no rational roots

Final result :

7a3 - 7a + 1

6 0

2 years ago

Read 2 more answers

Hell help me out thank youuuuuu byeee

Crank

Answer:

if im right its number 4!! if its not right sorry i did the math and thats what i got!!

4 0

2 years ago

How to solve 2x-y=17;x-y=10

Nataliya [291]

The answer is that y is equal to -3 and x is equal to 7
2x-y=17
-2(x-y)=10 )
--------------
2x-y=17
-2x+2y=-20
----------------
y=-3

x-y=10
x-(-3)=10
x+3=10
-3 -3
----------
x=7

8 0

3 years ago

Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).

I am Lyosha [343]

Take the homogeneous part and find the roots to the characteristic equation:

$y''+4y=0\implies r^2+4=0\implies r=\pm2i$

This means the characteristic solution is $y_c=C_1\cos2x+C_2\sin2x$ .

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form $y_p=ax\cos2x+bx\sin2x$ . Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With $y_1=\cos2x$ and $y_2=\sin2x$ , you're looking for a particular solution of the form $y_p=u_1y_1+u_2y_2$ . The functions $u_i$ satisfy

$u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx$

where $W(y_1,y_2)$ is the Wronskian determinant of the two characteristic solutions.

$W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2$

So you have

$u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx$
$u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x$

$u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx$
$u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x$

So you end up with a solution

$u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x$

but since $\cos2x$ is already accounted for in the characteristic solution, the particular solution is then

$y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x$

so that the general solution is

$y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x$

7 0

3 years ago