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2 years ago

Solve the equation -5x + 3 = 2x - 1

Mathematics

2 answers:

frozen [14]2 years ago

8 0

Step-by-step explanation:

Given

- 5x + 3 = 2x - 1

or, 2x + 5x = 3 + 1

or, 6x = 4

or, x = 4/ 6

Therefore x = 2 /3

Hope it will help :)❤

Verizon [17]2 years ago

3 0

Step-by-step explanation:

Subtract

from both sides of the equation

−

Simplify

Subtract the numbers

−

Subtract

from both sides of the equation

−

Simplify

Combine like terms

−

Divide both sides of the equation by the same term

−

Simplify

Cancel terms that are in both the numerator and denominator

Divide the numbers

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PLSSSS HELP!!!!

Taya2010 [7]

Answer:

The dilation is an enlargement by 3

Step-by-step explanation:

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7 0

2 years ago

Read 2 more answers

What is equivalent to the equation to 1/4 (8 + 6z + 12) ?

Fed [463]

1/4 (8 + 6z + 12)

------------------------------------------------------------
Combine like terms :
------------------------------------------------------------
1/4 (6z + 20)

------------------------------------------------------------
Apply distributive property :
------------------------------------------------------------
1/4(6z) + 1/4(20)
3/2z + 5

------------------------------------------------------------
Answer: 3/2z + 5
------------------------------------------------------------

7 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

Please help me on this!

Paladinen [302]

Answer:

The original value of the car.

Step-by-step explanation:

The 21,500 would be the <em>a </em>value of the function, as it is before the parentheses. The <em>a </em>value of an exponential function is a constant, and it is the y-intercept of that function, meaning in a real-world application, it would be where the values start at. In this case, the 21,500 is where the price of the car starts at, i.e., the original value of the car.

8 0

2 years ago

You have 1000000 small cubes.

Tju [1.3M]

Answer:

1000 in

Step-by-step explanation:

Given

The dimension of the small cube is 1\ in.1 in.

No of the cubes is 10^6106

If the small cubes are arranged on the floor

The area of the cubes is

\Rightarrow 10^6\times 1^2\ in.^2⇒106×12 in.2

8 0

2 years ago