Answer:
What is the price and costs of automobile tiresis?
Step-by-step explanation:
Answer:
Sum = 13+ 4i Difference = -5 - 8i
Step-by-step explanation:
Given complex number as (4 -2i) and (9 + 6i)
Sum of these complex number is
(4 -2i) + (9 + 6i) = ( 4 +9) + (6i -2i)
= 13 +4i
Difference of these complex number is
(4 -2i) - (9 + 6i) = (4 -9) - (2i +6i)
= -5 - 8i
Hence the sum and Difference of given complex numbers are
Sum = 13+ 4i Difference = -5 - 8i Answer
We know that<span>
<span>Figures can be proven similar if one, or more,
similarity transformations (reflections, translations, rotations, dilations)
can be found that map one figure onto another.
In this problem to prove circle 1 and circle 2 are similar, a
translation and a scale factor (from a dilation) will be found to map one
circle onto another.
we have that</span>
<span> Circle 1 is centered at (5,8) and has a
radius of 8 centimeters
Circle 2 is centered at (1,-2) and has a radius of 4 centimeters
</span>
step 1
<span>Move the center of the circle 1 onto the
center of the circle 2
the transformation has the following rule</span>
(x,y)--------> (x-4,y-10)
so
(5,8)------> (5-4,8-10)-----> (1,-2)
so
center circle 1 is now equal to center circle 2
<span>The circles are now concentric (they have the
same center)
</span>
step 2
<span>A dilation is needed to decrease the size of
circle 1 to coincide with circle 2
</span>
scale factor=radius circle 2/radius circle
1-----> 4/8----> 0.5
radius circle 1 will be=8*scale factor-----> 8*0.5-----> 4 cm
radius circle 1 is now equal
to radius circle 2
<span>A
translation, followed by a dilation will map one circle onto the other,
thus proving that the circles are similar
the answer is
</span></span>The circles are similar because you can translate Circle 1 using the transformation rule (x-4,y-10) and then dilate it using a scale factor of (0.5)
Let x and y be the upper and lower class limit of frequency distribution.
Given, width of the class = 5
⇒ x-y= 5 …
Also, given lower class (y) = 10 On putting y = 10, we get
x – 10= 5 ⇒ x = 15 So, the upper class limit of the lowest class is 15
Hence, the upper class limit of the highest class
=(Number of continuous classes x Class width + Lower class limit of the lowest class)
= 5 x 5+10 = 25+10=35
Hence,’the upper class limit of the highest class is 35.
<em><u>Alternate Method</u></em><em><u>.</u></em>
After finding the upper class limit of the lowest class, the five continuous classes in a frequency distribution with width 5 are 10-15,15-20, 20-25, 25-30 and 30-35.
Thus, the highest class is 30-35..
<h3>Hence, the upper limit of this class is 35.</h3>
<h2>_____________________</h2>
Answer:
Step-by-step explanation:
The trick is to find the third angle
180 - A - <BCA = <CBA
180 - X - <BCX = <CBX
Everything else is OK
<CBA = <CBX Equals equated to an equal express = an equal expression.
Now by a little slight of hand, you get the two triangles to be equal by ASA, which always works.
Cheating you say? There is no such thing as cheating if it correct and it works.
