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1 answer:
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Answer:
B'(16,14)
Step-by-step explanation:
In square ABCD, A(2,7) C(8,1) D(2,1). First, find the coordinates of the vertex B. From the attached diagram you can see that CD=6 units and is parallel to the x-axis, AD=6 units and is parallel to the y-axis. So, B(8,7).
The origin O is the center of dilation, 2 is the scale factor. The rule of the dilation is
(x,y)→(2x,2y)
Thus,
B(8,7)→B'(16,14) (see attached figure).
Answer and Step-by-step explanation:
Suppose that we have a number y which is a positive integer and that:
y =
Where;
= digit at 10⁰ => one's place (units place)
= digit at 10¹ => 10's place (tens place)
= digit at 10² => 100's place (hundreds place)
= digit at 10³ => 1000's place (thousands place)
.
.
.
= digit at 10ⁿ place
Then;
y = * 10⁰ +
* 10¹ +
* 10² +
* 10³ +
* 10⁴ +
* 10⁵ + . . . +
* 10ⁿ
<em>Since 10⁰ = 1, let's rewrite y as follows;</em>
y = +
* 10¹ +
* 10² +
* 10³ +
* 10⁴ +
* 10⁵ + . . . +
* 10ⁿ
Now, to test if y is divisible by 11, replace 10 in the equation above by -1. Since 10 divided by 11 gives -1 (mod 11) [mod means modulus]
y = +
* (-1)¹ +
* (-1)² +
* (-1)³ +
* (-1)⁴ +
* (-1)⁵ + . . . +
* (-1)ⁿ
=> y = -
+
-
+
-
+ . . . +
(-1)ⁿ (mod 11)
Therefore, it can be seen that, y is divisible by 11 if and only if alternating sum of its digits -
+
-
+
-
+ . . . +
(-1)ⁿ is divisible by 11
<em>Let's take an example</em>
Check if the following is divisible by 11.
i. 1859
<em>Solution</em>
1859 is divisible by 11 if and only if the alternating sum of its digit is divisible by 11. i.e if (1 - 8 + 5 - 9) is divisible by 11.
1 - 8 + 5 - 9 = -11.
Since -11 is divisible by 11 so is 1859
ii. 31415
<em>Solution</em>
31415 is divisible by 11 if and only if the alternating sum of its digit is divisible by 11. i.e if (3 - 1 + 4 - 1 + 5) is divisible by 11.
3 - 1 + 4 - 1 + 5 = 10.
Since 10 is not divisible by 11 so is 31415 not divisible.