Ask question

Ask question

All categories

2 years ago

5

Carlos walked to school on 14 of the 20 school days in February. Which value is equivalent to the fraction of the school days in

February that Carlos walked to
school?
A 70%
B. 0.07
C. 0.142
DO 56%

2 answers:

Leto [7]2 years ago

6 0

Answer: A

Step-by-step explanation: 14/20 = 70%

allsm [11]2 years ago

6 0

The answer is a - 70%

You might be interested in

Find a particular solution to the nonhomogeneous differential equation y??+4y?+5y=?10x+e^(?x).

Firdavs [7]

Answer:

A) Particular solution:

$2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

B) Homogeneous solution:

$y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))$

C) The most general solution is

$y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

Step-by-step explanation:

Given non homogeneous ODE is

$y''+4y'+5y=10x+e^{-x}---(1)$

To find homogeneous solution:

$D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)$

To find particular solution:

$y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\$

Substituting $y_{p},y'_{p},y''_{p}$ in (1)

$y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\$

Equating the coefficients

$5Ax+2Ce^{-x}+4A+5B=10x+e^{-x}\\5A=10\\A=2\\4A+5B=0\\B=-\frac{4A}{5}B=-\frac{8}{5}2C=1\\C=\frac{1}{2}\\So,\\y_{p}=2x+\frac{1}{2}e^{-x}-\frac{8}{5}---(3)\\$

The general solution is

$y=y_{h}+y_{p}$

from (2) ad (3)

$y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

6 0

2 years ago

The second statement is the of the first.

kipiarov [429]

B is the answer b.converse

5 0

1 year ago

Which expression gives the x-coordinate of the solution of the system 3x - y + z = 9?

Zolol [24]

×=y/3-z/3+3 thats the x value in this equation

8 0

2 years ago

The length of a 200 square foot rectangular vegetable garden is 4feet less than twice the width. Find the length and width of th

inna [77]

Answer:

Length = 18.099 ft

Width = 11.049 ft

Step-by-step explanation:

let the length of the field be x ft

and the width be y ft

as per the condition given in problem

x=2y-4 -----------(A)

Also the area is given as 200 sqft

Hence

xy=200

Hence from A we get

y(2y-4)=200

taking 2 as GCF out

2y(y-2)=200

Dividing both sides by 2 we get

$y(y-2)=100$

$y^2-2y=100$

subtracting 100 from both sides

$y^2-2y-100=0$

Now we solve the above equation with the help of Quadratic formula which is given in the image attached with this for any equation in form

$ax^2+bx+c=0$

Here in our case

a=1

b=-1

c=-100

Putting those values in the formula and solving them for y

$y=\frac{-(-2)+\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

Solving first

$y=\frac{2+\sqrt{4+400}{2}$

$y=\frac{2+\sqrt{404}{2}$

$y=\frac{2+20.099}{2}$

$y=\frac{22.099}{2}$

$y=11.049$

Solving second one

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{2-\sqrt{4+400}{2}$

$y=\frac{2-\sqrt{404}{2}$

$y=\frac{2-20.099}{2}$

$y=\frac{-18.99}{2}$

$y=-9.045$

Which is wrong as the width can not be in negative

Our width of the field is

y=11.099

Hence the length will be

x=2y-4

x=2(11.049)-4

x=22.099-4

x=18.099

Hence our length x and width y :

Length = 18.099 ft

Width = 11.049 ft

4 0

2 years ago

One fourth the sum of x and ten is identical to x minus 4."

Lynna [10]

Answer and Step-by-step explanation:

We are given the word format of an equation.

Let's right it out in numerical form.

$\frac{1}{4} (x + 10) = x - 4$

The first part says One fourth the sum of x and 10

The sum of x and 10 would be shown as x + 10, and one fourth of that means that $\frac{1}{4}$ is multiplying (x + 10).

In this case, identical would mean equal to.

So, $\frac{1}{4}$ (x + 10) would be = to something.

It says that it is identical to x minus 4 (x - 4).

$\frac{1}{4} (x + 10) = x - 4$ <u> is the equation.</u>

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

5 0

2 years ago