Is zero point 32 repeating times square root of 294 end square root rational or irrational? Justify your answer

Ludmilka [50]

Answer:

$the \: area \: of \: the \: logo \: is \: {91cm}^{2}$

Step-by-step explanation:

To determine the area of the logo you have to calculate the area of the triangle and the square that comform it and then add the four areas.

Area of the square.

To calculate the area of the square you have to calculate the square of one of its sides, following the formula:

$a = {a}^{2}$

Where "a" is the length of one of its side.

The side length of the square is a=7cm, so its area will be:

$asquare = {7}^{2} \\ asquare = {49cm}^{2}$

Area of the triangles.

The three triangles are equal, they have a base equal to the side of the square, i.e. with a length of 7cm, and their height is h=4cm.

To calculate the area of one triangle, you have to multiply its base by its height and divide by 2, following the formula:

$a = \frac{b.h}{2} \\ a = \frac{7.4}{2} \\ a = \frac{28}{2} \\ a = {14cm}^{2}$

The area calculated the correspond to one triangle, since all triangles are congruent, you have to multiply the said area by 3 to determine the area of three figures:

$atriangles = 3a \\ atriangles = {42cm}^{2}$

Now that the area of all shape are calculated, you have to add them to determine the area of the logo:

$alogo = asquare + atriangle \\ alogo = 49 + 42 \\ alogo = {91cm}^{2}$

4 0

3 years ago

Read 2 more answers

I need help please. If you are correct, I will give you Brainliest.

earnstyle [38]

Answer:

(C) 220

Step-by-step explanation:

Let x represent the number of adult tickets sold and y represent the number of student tickets sold. With the information given, we can set up two equations:

$x+y=360$

$5x+3y=1360$ (Since for every adult ticket sold, $5 is made and for every student ticket sold, $3 is made)

In the first equation, we can represent x in terms of y:

$x=360-y$

And then, we can substitute x in the second equation for 360 - y to get:

$5(360-y)+3y=1360$ which simplifies to:

$1800-2y=1360$ and therefore, $y=220$ .

Hope this helps :)

5 0

3 years ago

A scuba diver starts at 85 3/4 meters below the surface of the water and descends until he reaches 103 1/5

RSB [31]

Answer:

From his starting point of 85 3/4, he descended 17 9/20 meters

Step-by-step explanation:

We want to know how much farther he went down from his starting point.

103 1/5 - 85 3/4

S

tart off by making the denominators of both the numbers the same.

103 1/5 - 85 3/4

103 4/20 - 85 15/20

Make them into improper fractions

2064/20 - 1715/20 = 349/20

Make 349/20 into a mixed fraction

17 9/20

5 0

3 years ago

Consider the following equation of the form dy/dt = f(y)dy/dt = ey − 1, −[infinity] < y0 < [infinity](a) Sketch the graph

kotegsom [21]

Complete Question:

The complete question is shown on the first uploaded image

Answer:

a) The graph of f(y) versus y. is shown on the second uploaded image

b) The critical point is at y = 0 and the solution is asymptotically unstable.

c)The phase line is shown on the third uploaded image

d) The sketch for the several graphs of solution in the ty-plane is shown on the fourth uploaded image

Step-by-step explanation:

Step One: Sketch The Graph of f(y) versus y

Looking at the given differential equation

$\frac{dy}{dt} = e^{y} - 1$ for -∞ < $y_{o}$ < ∞

We can say let $\frac{dy}{dt}$ = f(y) = $e^{y} - 1$

Now the dependent value is f(y) and the independent value is y so to sketch is graph we can assume a scale in this case i cm on the graph is equal to 2 unit for both f(y) and y and the match the coordinates and after that join the point to form the graph as shown on the uploaded image.

Step Two : Determine the critical point

To fin the critical point we have to set $\frac{dy}{dt}$ = 0

This means $e^{y} - 1$ = 0

For this to be possible $e^{y}$ = 1

which means that $e^{y}$ = $e^{0}$

which implies that y = 0

Hence the critical point occurs at y = 0

meaning that the equilibrium solution is y = 0

As t → ∞, our curve is going to move away from y = 0 hence it is asymptotically unstable.

Step Three : Draw the Phase lines

A phase line can be defined as an image that shows or represents the way an ODE(ordinary differential equation ) that does not explicitly depend on the independent variable behaves in a single variable. To draw this phase line , draw the y-axis as a vertical line and mark on it the equilibrium, i.e. where f(y) = 0.

In each of the intervals bounded by the equilibrium draw an upward

pointing arrow if f(y) > 0 and a downward pointing arrow if f(y) < 0.

This phase line would solely depend on y does not matter what t is

On the positive x axis it would get steeper very quickly as you move up (looking at the part A graph).

For below the x-axis which stable (looking at the part a graph) we are still going to have negative slope but they are going to be close to 0 and they would take a little bit longer to get steeper

Step Four : Draw a Solution Curve

A solution curve is a curve that shows the solution of a DE (deferential equation)

Here the solution curve would be drawn on the ty-plane

So the t-axis(x-axis) is its the equilibrium that is it is the solution

If we are above the x-axis it is going to increase faster and if we are below it is going to decrease but it would be slower (looking at part A graph)

5 0

3 years ago

Ms. Ache is paid $1250 per week but is fined $100 each day she is late to work. Ms. Ache wants to make at least $3,000 over the

faust18 [17]

She can have 7 days off at most because 3 X 1250 = 3750 - 700 = 3050

7 0

3 years ago