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3 years ago

What's -3 1/8 divided by -1 2/3?

2 answers:

8 0

(-3 1/8) / (- 1 2/3)....turn them to improper fractions
(- 25/8) / (- 5/3) ..when dividing fractions, " flip " what u r dividing by, then multiply
-25/8 * - 3/5 =
-75/-40 which reduces to 15/8 or 1 7/8

choli [55]3 years ago

4 0

1.875 is what you are looking for.

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Which equation represents the graph of Charli's line? A. y= -3/4x-3. B. Y= 3/4x- 4. C. y = -4/3x- 4. D. y= 4/3x- 3. LOOK AT THE

enot [183]

Saya rase jwpn di C . Kalau bukan minx maaf lah ye

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3 years ago

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If x = 5, y = 2 and z = 3 then xz (4y - 2z) is

WARRIOR [948]

(4x2-2x3)
(8-6)
xz=2

Hope this helps!

4 0

3 years ago

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What is the value of x?

Nady [450]

Answer: X equals B.

Step-by-step explanation: all the degrees in a triangle add to be 180, so add up the given angles and subtract from 180 to get the missing angle.

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3 years ago

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Adena, Julius, and Tia volunteered to read to children at the public library. Julius worked two hours less than Tia. Adena worke

mafiozo [28]

Answer:

28 hours

Step-by-step explanation:

Adena, Julius, and Tia volunteered to read to children at the public library.

Let us represent, the number of hours worked by

Adena = x

Julius = y

Tia = z

Julius worked two hours less than Tia.

y = z - 2

z = y + 2

Adena worked twice as many hours as Julius.

x = 2y

Altogether they worked 58 hours.

x + y + z = 58.... Equation 1

We substitute 2y for x and y + 2 for z in Equation 1

2y + y + y + 2 = 58

4y + 2 = 58

Collect like terms

4y = 58 - 2

4y = 56

y = 56/4

y = 14

We are the find the number of hours Adena worked which is represented by x

Note that:

x = 2y

y = 14, hence,

x = 2 × 14

x = 28 hours

Therefore, the number of hours Adena worked is 28 hours

4 0

3 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago