Ask question

Ask question

All categories

Ira Lisetskai [31]

3 years ago

14

Suppose point C is at (–1, –1). Use the numbers below to explain the distance that a fourth point, D, should be from the existin

g points to form a square. Numbers may be used once, more than once, or not at all.

1 answer:

Leno4ka [110]3 years ago

5 0

Answer:

whats the anwserrrrrrrr

Step-by-step explanation:

You might be interested in

Write an equation that is parallel to y=-1/4x+2

AlexFokin [52]

Answer:

y = -1/4 -5

Step-by-step explanation:

Any equation will be parallel to your base equation as long as the y intercept is different. To write a parallel equation, you just have to change the y intercept.

For example, I changed it from 2 to negative 5.

5 0

3 years ago

PLS PLS PLS HELP ME WITH THIS!!!!!!

nlexa [21]

Answer:

720

Step-by-step explanation:

8 0

3 years ago

Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product o

Keith_Richards [23]

Answer:

32.1 g

Step-by-step explanation:

In each 3 grams of C, there are 2 grams of A and 1 gram of B. So, for some amount C, the amount remaining of A is 40 -(2C/3), and the amount remaining of B is (50 -C/3). Since the reaction rate is proportional to the product of these amounts, we have ...

C' = k(40 -2C/3)(50 -C/3) = (2k/9)(60 -C)(150 -C) . . . for some constant k

This is separable differential equation with a solution of the form ...

ln((150 -C)/(60 -C)) = at + b

We know that C(0) = 0, so b=ln(150/60) = ln(2.5). And we know that C(10) = 20, so ln(130/40) = 10a +ln(2.5) ⇒ a = ln(1.3)/10

Then our equation for C is ...

ln((150 -C)/(60 -C)) = t·ln(1.3)/10 +ln(2.5)

__

For t=20, this is ...

ln((150 -C)/(60 -C)) = 2ln(1.3) +ln(2.5) = ln(2.5·1.3²) = ln(4.225)

Taking antilogs, we have ...

(150 -C)/(60 -C) = 4.225

1 +90/(60 -C) = 4.225

C = 60 -90/3.225 ≈ 32.093 . . . . . grams of product in 20 minutes

In 20 minutes, about 32.1 grams of C are formed.

7 0

3 years ago

Find a particular solution to y" - y + y = 2 sin(3x)

leonid [27]

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

y"=z'

$\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x$

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

z'+z(x-1)=2 sin 3 x-----------(1)

This is a type of linear differential equation.

Integrating factor

$=e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}$

Multiplying both sides of equation (1) by integrating factor and integrating we get

$\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I$

$I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx$

8 0

3 years ago

3 to the 4th power + 2.5

Marina CMI [18]

Answer:

83.5

Step-by-step explanation:

$3^4+2.5\\81+2.5\\83.5\\$

7 0

3 years ago