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4 years ago

Which of these numbers is irrational (The demontar on the last fraction is 3 BTW)

Mathematics

2 answers:

Sedbober [7]4 years ago

7 0

Letter a would be the best answer

nataly862011 [7]4 years ago

3 0

An irrational number can be written as a decimal, but not as a fraction. An irrational number has endless non-repeating digits to the right of the decimal point. The answer is A. Please mark brainliest

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Can someone please help?!

Bond [772]

Answer:

1. $R_{90} (x,y)=(-y,x)$

2. $R_{180} (x,y)=(-x,-y)$

3. $R_{270} (x,y)=(y,-x)$ .

Step-by-step explanation:

Let us assume a point (x,y) in the co-ordinate plane on which the transformations will be applied.

Now, we know that 'rotation' is a transformation that turns that image to a certain degree about a point.

So, the given transformations gives us the forms as:

1. When we rotate an ( x,y ) by 90° about origin counter-clockwise, the resultant co-ordinate is ( -y,x ).

So, the function form is $R_{90} (x,y)=(-y,x)$ .

2. When we rotate an ( x,y ) by 180° about origin counter-clockwise, the resultant co-ordinate is ( -x,-y ).

So, the function form is $R_{180} (x,y)=(-x,-y)$ .

3. When we rotate an ( x,y ) by 270° about origin counter-clockwise, the resultant co-ordinate is ( y,-x ).

So, the function form is $R_{270} (x,y)=(y,-x)$ .

7 0

3 years ago

Suppose that we agree to pay you 9¢ for every problem in this chapter that you solve correctly and fine you 5¢ for every probl

Mariulka [41]

The answer is 10 problems. Hope it helps

4 0

3 years ago

Find the area of the surface generated by revolving x=t + sqrt 2, y= (t^2)/2 + sqrt 2t+1, -sqrt 2 <= t <= sqrt about the y

EleoNora [17]

The area is given by the integral

$\displaystyle A=2\pi\int_Cx(t)\,\mathrm ds$

where <em>C</em> is the curve and $dS$ is the line element,

$\mathrm ds=\sqrt{\left(\dfrac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt$

We have

$x(t)=t+\sqrt 2\implies\dfrac{\mathrm dx}{\mathrm dt}=1$

$y(t)=\dfrac{t^2}2+\sqrt 2\,t+1\implies\dfrac{\mathrm dy}{\mathrm dt}=t+\sqrt 2$

$\implies\mathrm ds=\sqrt{1^2+(t+\sqrt2)^2}\,\mathrm dt=\sqrt{t^2+2\sqrt2\,t+3}\,\mathrm dt$

So the area is

$\displaystyle A=2\pi\int_{-\sqrt2}^{\sqrt2}(t+\sqrt 2)\sqrt{t^2+2\sqrt 2\,t+3}\,\mathrm dt$

Substitute $u=t^2+2\sqrt2\,t+3$ and $\mathrm du=(2t+2\sqrt 2)\,\mathrm dt$ :

$\displaystyle A=\pi\int_1^9\sqrt u\,\mathrm du=\frac{2\pi}3u^{3/2}\bigg|_1^9=\frac{52\pi}3$

8 0

3 years ago

If square ABCD is defined by the coordinates A(0, 0), B(0, -5), C(-5, -5), D(-5, 0) is dilated by a scale factor of 2, with resu

ruslelena [56]

To obtain the center of dilation we use the formula:
IOA'I/IOAI=IfI
this can be written as:
IOA'I=IOAIIfI
where O is the center of dilation; suppose our center is (x,y) thus plugging our values we get:
√(5-x)²+(5-y)²=2[√(0-x)²+(0-y)²]
√(5-x)²+(5-y)²=2√(x²+y²)
squaring both sides we get:
(5-x)²+(5-y)²=4(x²+y²)
to solve the above we equate as follows:
(5-x)²=4x²
x=-5 or 5/3
also
(5-y)²=4y²
y=-5 or 5/3
thus the center of dilation is:
(-5,-5)

5 0

3 years ago

Read 2 more answers

Answer the Question on the picture answer part C

Sveta_85 [38]

Answer:

Have you done the previous parts? I need the answer to those to do C

3 0

3 years ago