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3 years ago

What is the solution to this equation? X/4 =-12

Mathematics

1 answer:

Natasha2012 [34]3 years ago

7 0

Answer:

C. x=-48

Step-by-step explanation:

x/4=-12

multiply both sides by 4

4x/4=4 (-12)

simplify

x=-48

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Step-by-step explanation:

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2 years ago

Madeline has used 50 sheets of a roll of paper towels, which is 5/8 of the entire roll. The equation 5/8s = 50 can be used to fi

raketka [301]

Answer:

s=80

Step-by-step explanation:

5/8ths is .625

80 x .625 =50

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3 years ago

Find the vertex of " "y = x^2 + 5x+ 4" please show steps as well thanks

zlopas [31]

(x^2 +5x+2.5^2) +4+5

(x+2.5)^2+9

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8 0

2 years ago

Read 2 more answers

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

Can someone explain? And answer?

alexandr402 [8]

Answer:

$25.00

Step-by-step explanation:

You already know that $85 is an additional fee added to the monthly fees.

So, you first have to subtract 85 from $685 to get the total amount of monthly fees paid over 2 years.

685-85= 600

So, you can conclude that $600 is the total amount of monthly fees paid.

Now, you want to divide 600 by (the number of months in 2 years) to get the amount of monthly fees.

Since there is 24 months in 2 years you do:

600/ 24 which equals 25.

In conclusion, you can find that the monthly fee is $25.00.

6 0

3 years ago