What’s the square root of -100

What is the sum of -4 1/2 + 3 14

Evgen [1.6K]

Answer:

-1 1/4

Step-by-step explanation:

7 0

3 years ago

Will give brainiest Find the equation of the straight line parallel to the y-axis and passes through the point (-2, 3). _____

IrinaVladis [17]

Answer:

Step-by-step explanation:

8 0

2 years ago

Coefficient of b in expansion of (3+b)^4

Margarita [4]

Are you sure you want ONLY the coefficient of b? If you expand this, you will have b in 3 of 4 terms.

According to Pascal's Triangle, the coefficients of (a+b)^4 are as follows:

1
1 2 1
1 3 3 1
1 4 6 4 1

So (a+b)^4 would be 1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

Here, you want (3 + b)^4. Here's what that looks like:

3^4 + 4[3^3*b] + 6[3^2*b^2] + 4[3*b^3] + 1[b^4]

Which coeff did you want?

4 0

3 years ago

Let FS R be a finite set. (a) Prove that F is closed (b) Prove that every point in F is an isolated point.

Lorico [155]

Answer:

a) Suppose that F is ordered in ascending order: $F = \{x_1,\ldots, x_n\}$ . Then, the complement of F can be written as

$F^c = (-\infty,x_1)\cup (x_1,x_2)\cup (x_2,x_3)\cup \cdots \cup (x_{n-1}, x_{n})\cup (x_n,+\infty)$

which is the union of a finite number of open intervals, then $F^c$ is an open set. Thus, F is a closed subset of the real numbers.

b) Take an arbitrary element of F, let us say $x_k$ . Now, choose a real number $\epsilon$ such that

$0$ there are not other element of F, because $\epsilon$ is less that the minimum distance between $x_k$ and its neighbors.

In case that $k=1$ we only consider $0$ , and if $k=n$ we only consider $0$ .

Then, all points of F are isolated.

Step-by-step explanation:

8 0

3 years ago

A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th

Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let $S(t)$ be the amount of sugar in the tank at time $t$ . Then $S(0)=25$ .

Sugar is added to the tank at a rate of P lb/min, and removed at a rate of

$\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}$

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

$\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}$

Separate variables, integrate, and solve for S.

$\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt$

$\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt$

$-100\ln\left|P-\dfrac S{100}\right|=t+C$

$\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t$

$P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}$

$\dfrac S{100}=P-Ce^{-100t}$

$S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}$

Use the initial value to solve for C :

$S(0)=25\implies 25=100P-C\implies C=100P-25$

$\implies S(t)=100P-(100P-25)e^{-100t}$

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

$1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}$

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick P such that

$S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}$

5 0

3 years ago