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3 years ago

How do change 0.9 in fraction

Mathematics

2 answers:

yarga [219]3 years ago

6 0

You change it to 9/10

Fantom [35]3 years ago

3 0

0.9 as a fraction is: $\frac{9}{10}$

What place is the '9' in? It's in the tenth place.
0.9 can be said as $nine \ tenths$

Nine $(9)$ tenths $(10)$ = $\frac{9}{10}$

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Sketch the region that is common to the graphs of x ≥ 2,

Oliga [24]

a. Find the graph of their common region in the attachment

b. The area of the common region of the graphs is 8 units²

<h3>a. How to sketch the region common to the graphs?</h3>

Since we have x ≥ 2, y ≥ 0, and x + y ≤ 6, we plot each graph separately and find their region of intersection.

The graph of x ≥ 2 is the region right of the line x = 2.

The graph of y ≥ 0 is the region above the line y = 0 or x-axis.

To plot the graph of x + y ≤ 6, we first plot the graph of x + y = 6 ⇒ y = -x + 6. Then the graph of x + y ≤ 6 is the graph of y ≤ - x + 6.

So, the graph of x + y ≤ 6 is the region below the line y = - x + 6

From the graph, the regions intersect at (2, 0), (2, 4) and (6, 0)

Find the graph of their common region in the attachment

<h3>b. The area of the common region</h3>

From the graph, we see that the common region is a right angled triangle with

height = 4 units and
base = 4 units

So, its area = 1/2 × height × base

= 1/2 × 4 units × 4 units

= 1/2 × 16 units²

= 8 units²

So, the area of the common region of the graphs is 8 units²

Learn more about region common to graphs here:

brainly.com/question/27932405

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2 years ago

Which undefined geometric term is described as a two-dimensional set of points that has no beginning or end?

IgorC [24]

Plane, it contains tons of 2d points with no set beginning or end

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3 years ago

A trough of water is 20 meters in length and its ends are in the shape of an isosceles triangle whose width is 7 meters and heig

Vaselesa [24]

Answer:

a) Depth changing rate of change is $0.24m/min$ , When the water is 6 meters deep

b) The width of the top of the water is changing at a rate of $0.17m/min$ , When the water is 6 meters deep

Step-by-step explanation:

As we can see in the attachment part II, there are similar triangles, so we have the following relation between them $\frac{3.5}{10} =\frac{a}{h}$ , then $a=0.35h$ .

a) As we have that volume is $V=\frac{1}{2} 2ahL=ahL$ , then $V=(0.35h^{2})L$ , so we can derivate it $\frac{dV}{dt}=2(0.35h)L\frac{dh}{dt}$ due to the chain rule, then we clean this expression for $\frac{dh}{dt}=\frac{1}{0.7hL}\frac{dV}{dt}$ and compute with the knowns $\frac{dh}{dt}=\frac{1}{0.7(6m)(20m)}2m^{3}/min=0.24m/min$ , is the depth changing rate of change when the water is 6 meters deep.

b) As the width of the top is $2a=0.7h$ , we can derivate it and obtain $\frac{da}{dt}=0.7\frac{dh}{dt} =0.7*0.24m/min=0.17m/min$ The width of the top of the water is changing, When the water is 6 meters deep at this rate

8 0

3 years ago

One angle of a parallelogram measures 110°. What are the measures of the other three angles in the parallelogram?

Art [367]

Answer:

70, 110, 70

Step-by-step explanation:

The measures of the adjacent angles of a parallelogram always add up to be 180 degrees (they are supplementary).

4 0

3 years ago

Sketch a graph of the polynomial function f(x) =x3− 2x2. Use it to complete the following:

AlladinOne [14]

Answer:

We have the function:

f(x) = x^3 - 2*x^2

To sketch this, we need to graph some points, and then just draw a line that passes through the points.

The graph of this equation is shown below.

Now we can complete the question.

If the graph is below the x-axis in some interval, the function is negative in that interval

If the graph is above the x-axis in some interval, the function is positive in that interval.

If the graph goes up in a interval, then the function is increasing in that interval

If the graph goes down on an interval, then the function is decreasing in that interval.

Then:

1) f is------ on the intervals (−∞, 0) and (0, 2).

Here we can see that the graph is below the x-axis in those intervals, then here we have:

f is negative on the intervals (−∞, 0) and (0, 2).

2) f is------ on the interval (2,∞)

Here the answer is positive:

f is positive on the interval (2,∞)

3) fi is ------ on the interval (0, 4/3)

In the graph, you can see that the graph goes down in that interval, then the correct answer here is:

f is decreasing on the interval (0, 4/3)

4) f is------ on the intervals (−∞, 0) and (4/3, ∞).

In this case, we can see that the graph goes up in these intervals, then the correct answer here is:

f is increasing on the intervals (−∞, 0) and (4/3, ∞).

5 0

3 years ago