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3 years ago

How do I solve and simplify this if needed?

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

7 0

1 1/2 divided by 2 1/4
Turn them into improper fractions
3/2 divided by 9/4
Change the problem to 3/2 x 4/9
You can cross simplify (1 x 2/3 = 2/3)
Or you can multiply straight across:
3/2 x 4/9 = 12/18
And simplify (12/6 and 18/6)
2/3

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What is the value of x if e3x+6=8

goldfiish [28.3K]

Answer:

x = 1/3 ln(2)

Step-by-step explanation:

e^(3x)+6=8

Subtract 6 from each side

e^(3x)+6-6=8-6

e^(3x) = 2

Take the natural log of each side

ln (e ^3x) = ln (2)

3x = ln(2)

Divide by 3

3x/3 = 1/3 ln(2)

x = 1/3 ln(2)

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4 years ago

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Determine the equation of the line, written in slope-intercept form, that passes through the points (1,1)

nataly862011 [7]

Answer:

y=-1/2x+3/2

Step-by-step explanation:

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3 years ago

X + 3 - 3=-6 show all work

docker41 [41]

Answer:

x = -6

Step-by-step explanation:

$x+3-3=-6~(Given)\\x=-6~(Combine~like~terms)$

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3 years ago

Solve 2(x-3)>-3(-3+x) (Picture added, multiple choice)

Rudiy27

Answer:

Answer: D

Step-by-step explanation:

$2(x - 3) \geqslant - 3( - 3 + x)$

open the brackets [ distributive property ]

$2x - 6 \geqslant 9 - 3x$

collect like terms:

$2x + 3x \geqslant 9 + 6 \\ 5x \geqslant 15$

divide either sides by 5:

$\frac{5x}{5} \geqslant \frac{15}{5} \\ \\ { \underline{ \bf{ \: \: x \geqslant 3 \: \: }}}$

3 0

3 years ago

Two solutions to y'' – 2y' – 35y = 0 are yı = e, Y2 = e -5t a) Find the Wronskian. W = 0 Preview b) Find the solution satisfying

pashok25 [27]

Answer:

a. $w(t)=-12e^{2t}$

b. $y(t)=-\frac{9}{2}e^{7t}-\frac{5}{2}e^{-5t}$

Step-by-step explanation:

We have a differential equation

y''-2 y'-35 y=0

Auxillary equation

$(D^2-2D-35)=0$

By factorization method we are finding the solution

$D^2-7D+5D-35=0$

$(D-7)(D+5)=0$

Substitute each factor equal to zero

D-7=0 and D+5=0

D=7 and D=-5

Therefore ,

General solution is

$y(x)=C_1e^{7t}+C_2e^{-5t}$

Let $y_1=e^{7t} \;and \;y_2=e^{-5t}$

We have to find Wronskian

$w(t)=\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}$

Substitute values then we get

$w(t)=\begin{vmatrix}e^{7t}&e^{-5t}\\7e^{7t}&-5e^{-5t}\end{vmatrix}$

$w(t)=-5e^{7t}\cdot e^{-5t}-7e^{7t}\cdot e^{-5t}=-5e^{7t-5t}-7e^[7t-5t}$

$w(t)=-5e^{2t}-7e^{2t}=-12e^{2t}$

a. $w(t)=-12e^{2t}$

We are given that y(0)=-7 and y'(0)=23

Substitute the value in general solution the we get

$y(0)=C_1+C_2$

$C_1+C_2=-7$ ....(equation I)

$y'(t)=7C_1e^{7t}-5C_2e^{-5t}$

$y'(0)=7C_1-5C_2$

$7C_1-5C_2=23$ ......(equation II)

Equation I is multiply by 5 then we subtract equation II from equation I

Using elimination method we eliminate $C_1$

Then we get $C_2=-\frac{5}{2}$

Substitute the value of $C_2$ in I equation then we get

$C_1-\frac{5}{2}=-7$

$C_1=-7+\frac{5}{2}=\frac{-14+5}{2}=-\frac{9}{2}$

Hence, the general solution is

b. $y(t)=-\frac{9}{2}e^{7t}-\frac{5}{2}e^{-5t}$

7 0

3 years ago