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Wittaler [7]
3 years ago
13

5x-y>4 how to solve this one​

Mathematics
1 answer:
Allushta [10]3 years ago
3 0

If you want me to solve for "Y" then, the answer will be,

y  < 5x - 4

ir if you want me to solve for "X" instead then the answer will be,

x >  \frac{4 + y}{5}

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your sister can get $30 per day for selling clothes at the market. but she has to $60 at first for renting a stand. you can get
ella [17]
Day Sis you
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8 0
3 years ago
Read 2 more answers
Please answer ASAP!
7nadin3 [17]

Answer:

h = 4.5 inches

Step-by-step explanation:

The area of a trapezoid is given by

A = 1/2 (b1+b2) h

We know A b1 and b2

20.25 = 1/2 (5+4)*h

20.25 = 9/2 h

Multiply each side by 2/9 to isolate h

2/9 * 20.25 = 2/9 *9/2 h

4.5 =h

6 0
4 years ago
x = c1 cos(t) + c2 sin(t) is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the seco
igomit [66]

Answer:

x=-cos(t)+2sin(t)

Step-by-step explanation:

The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0

Will have a characteristic equation of the form:

a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0

Where solutions r_1,r_2...,r_n are the roots from which the general solution can be found.

For real roots the solution is given by:

y(t)=c_1e^{r_1t} +c_2e^{r_2t}

For real repeated roots the solution is given by:

y(t)=c_1e^{rt} +c_2te^{rt}

For complex roots the solution is given by:

y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)

Where:

r_1_,_2=\lambda \pm \mu i

Let's find the solution for x''+x=0 using the previous information:

The characteristic equation is:

r^{2} +1=0

So, the roots are given by:

r_1_,_2=0\pm \sqrt{-1} =\pm i

Therefore, the solution is:

x(t)=c_1cos(t)+c_2sin(t)

As you can see, is the same solution provided by the problem.

Moving on, let's find the derivative of x(t) in order to find the constants c_1 and c_2:

x'(t)=-c_1sin(t)+c_2cos(t)

Evaluating the initial conditions:

x(0)=-1\\\\-1=c_1cos(0)+c_2sin(0)\\\\-1=c_1

And

x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2

Now we have found the value of the constants, the solution of the second-order IVP is:

x=-cos(t)+2sin(t)

3 0
3 years ago
Someone check my answers please
Anna007 [38]

Answer:

its right

Step-by-step explanation:

4 0
3 years ago
Simplify<br><br> Will mark brainliest, please show work if possible
Reil [10]
The answer for this question is 8/1

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3 years ago
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