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3 years ago

How many nanoseconds does it take for a computer to perform one calculation if it performs 6.7 × 10^{7} calculations per second?

Mathematics

1 answer:

crimeas [40]3 years ago

8 0

Answer:

the comuter does 1 calculation in 1.5* $10^{-17}$ nanosec

Step-by-step explanation:

First step: we need to write all terms in nanosec, as we know:

1[nanosec]= $10^{-9}$ [sec]

The computer does 6.7* $10^{7}$ [ $\frac{calculation}{sec}$ ]

if we multiplicate for $\frac{sec}{nanosec*10^{*9}}$ , from the units convertion.

6.7* $10^{7}$ [ $\frac{calculation}{sec}$ * $\frac{sec}{nanosec*10^{*9}}$

6.7* $10^{16}$ [ $\frac{calculation}{nanosec}$ ]

Now doing a simple three rule we have this:

6.7 * $10^{16}$ calculation in ⇒ 1 nanosec

1 calculation in ⇒ $\frac{1 nanosec* 1calculation}{6.7*10^{16} calculation }$

So, we found: the comuter does 1 calculation in 1.5* $10^{-17}$ nanosec

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If m FDG = 50°, what is m FEG? 75° 50° 100° 25° Thank you!

denpristay [2]

The measure of the angle FEG is probably the same as the measure of the angle FDG. There is nothing saying that EF is parallel to DG, but if so, the measure of the angle FEG is also 50º.
When the diagonals of a trapezium with two parallel bases inside of a circle are drawn, they make the same angle measure with those bases.

7 0

3 years ago

Find the domain for the function f(x)=sqrt x^2-x+6

dybincka [34]

Answer:

The domain is {x : x ∈ R} or (-∞ , ∞)

Step-by-step explanation:

* Lets explain how to find the domain

- The domain of the function is the values of x which make the

function defined

- The quantity under the square root must be ≥ 0 because there is

no square root for negative value

* Lets solve the problem

∵ f(x) = √(x² - x + 6)

∴ The value of (x² - x + 6) must be greater than or equal zero because

there is no square root for negative value

- Graph the function to know which values of x make the quantity

under the root is negative that means the values of x which make

the graph under the x-axis

∵ The graph doesn't intersect the x-axis at any point

∵ All the graph is above the x-axis

∴ There is no value of x make f(x) < 0

∴ x can be any real number

∴ The domain of f(x) is all real numbers

∴ The domain is {x : x ∈ R} or (-∞ , ∞)

6 0

3 years ago

How do you solve: 5x^2 + 25x - 70

Oxana [17]

Equation at the end of step 1 : (5x2 - 25x) - 70 = 0 Step 2 :Step 3 :Pulling out like terms :

3.1 Pull out like factors :

 5x2 - 25x - 70 = 5 • (x2 - 5x - 14)

Trying to factor by splitting the middle term

3.2 Factoring x2 - 5x - 14

The first term is, x2 its coefficient is 1 .
The middle term is, -5x its coefficient is -5 .
The last term, "the constant", is -14 

Step-1 : Multiply the coefficient of the first term by the constant 1 • -14 = -14

Step-2 : Find two factors of -14 whose sum equals the coefficient of the middle term, which is -5 .

-14 + 1 = -13 -7 + 2 = -5 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -7 and 2
 x2 - 7x + 2x - 14

Step-4 : Add up the first 2 terms, pulling out like factors :
 x • (x-7)
 Add up the last 2 terms, pulling out common factors :
 2 • (x-7)
Step-5 : Add up the four terms of step 4 :
 (x+2) • (x-7)
 Which is the desired factorization

Equation at the end of step 3 : 5 • (x + 2) • (x - 7) = 0 Step 4 :Theory - Roots of a product :

4.1 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

Equations which are never true :

4.2 Solve : 5 = 0

This equation has no solution.
A a non-zero constant never equals zero.

Solving a Single Variable Equation :

4.3 Solve : x+2 = 0

Subtract 2 from both sides of the equation :
 x = -2

Solving a Single Variable Equation :

4.4 Solve : x-7 = 0

Add 7 to both sides of the equation :
 x = 7

Supplement : Solving Quadratic Equation DirectlySolving x2-5x-14 = 0 directly

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Parabola, Finding the Vertex :

5.1 Find the Vertex of y = x2-5x-14

Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting "y" because the coefficient of the first term, 1 , is positive (greater than zero).

Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.

Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.

For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is 2.5000 

Plugging into the parabola formula 2.5000 for x we can calculate the y -coordinate :
 y = 1.0 * 2.50 * 2.50 - 5.0 * 2.50 - 14.0
or y = -20.250

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = x2-5x-14
Axis of Symmetry (dashed) {x}={ 2.50}
Vertex at {x,y} = { 2.50,-20.25}
x -Intercepts (Roots) :
Root 1 at {x,y} = {-2.00, 0.00}
Root 2 at {x,y} = { 7.00, 0.00}

Solve Quadratic Equation by Completing The Square

5.2 Solving x2-5x-14 = 0 by Completing The Square .

Add 14 to both side of the equation :
 x2-5x = 14

Now the clever bit: Take the coefficient of x , which is 5 , divide by two, giving 5/2 , and finally square it giving 25/4

Add 25/4 to both sides of the equation :
 On the right hand side we have :
 14 + 25/4 or, (14/1)+(25/4)
 The common denominator of the two fractions is 4 Adding (56/4)+(25/4) gives 81/4
 So adding to both sides we finally get :
 x2-5x+(25/4) = 81/4

Adding 25/4 has completed the left hand side into a perfect square :
 x2-5x+(25/4) =
 (x-(5/2)) • (x-(5/2)) =
 (x-(5/2))2 
Things which are equal to the same thing are also equal to one another. Since
 x2-5x+(25/4) = 81/4 and
 x2-5x+(25/4) = (x-(5/2))2 
then, according to the law of transitivity,
 (x-(5/2))2 = 81/4

We'll refer to this Equation as Eq. #5.2.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
 (x-(5/2))2 is
 (x-(5/2))2/2 =
 (x-(5/2))1 =
 x-(5/2)

Now, applying the Square Root Principle to Eq. #5.2.1 we get:
 x-(5/2) = √ 81/4 

Add 5/2 to both sides to obtain:
 x = 5/2 + √ 81/4 

Since a square root has two values, one positive and the other negative
 x2 - 5x - 14 = 0
 has two solutions:
 x = 5/2 + √ 81/4 
 or
 x = 5/2 - √ 81/4 

Note that √ 81/4 can be written as
 √ 81 / √ 4 which is 9 / 2