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deff fn [24]
3 years ago
10

On a coordinate plane, line segment A prime B has points (negative 2, negative 4) and (2, 1). Find the pre-image given the image

and scale factor.     Scale factor: n = 2 Identify the coordinates of the image.    A'(–2, –4)   B'(2, 1) Divide the coordinates by the scale factor. The pre-image of B' is B(1, 1 2 ). What is the pre-image of A'?
Mathematics
2 answers:
Burka [1]3 years ago
8 0

Answer:

-1,-2

Step-by-step explanation:

i guessed and got it right

GenaCL600 [577]3 years ago
7 0

Answer:The answer is -1,-2

Step-by-step explanation:TOOK TEST ON EDG

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PLS HELP ME WITH THIS QUESTION!!!!!
sergiy2304 [10]

Answer:

D is the answer

Step-by-step explanation:

4 0
2 years ago
How many 5-digit palindromes contain only even digits?
34kurt

Answer:

The correct answer is - 900

Step-by-step explanation:

A palindrome is a positive integer that is same whether read from left to right or from right to left. For example, 123321 palindromes.

An n-digit palindrome is determined from the first  n/2  digits if n is even, and from the first n+1/2 digits if n is odd.

Therefore, if  n  is even, there are  9×10⁽n⁻²/²⁾  palindromes; and if n is odd, there are  9×10⁽n₋²/²⁾ palindromes.  where n is number of digits.

For  n=5 , there are  9×10²=900 palindromes

3 0
3 years ago
What is 58,000,000,000 estimated as the product of a single digit and a power of 10?
brilliants [131]
The\ scientific\ notation:a\cdot10^n\ where\ 1\leq a \ \textless \  10\ and\ n\in\mathbb{Z}.\\\\5\underbrace{8,000,000,000}_{\leftarrow10}=5.8\cdot10^{10}\approx6\cdot10^{10}\to\fbox{D.}
5 0
3 years ago
Read 2 more answers
Find the area of the triangle<br> 1. C=110 degrees, a=6, b=10<br> 2. B=130 degrees, a=92, c=30
wel
1.\ \ \ |\angle \ ACB|=110^0\\\\.\ \ \ \ \ a=|BC|=6\ [u]\ \ \ and\ \ \ b=|AC|=10\ [u]\\\\Area= \frac{1}{2} \cdot |AC|\cdot |BC|\cdot sin(|\angle \ ACB|)\\\\Area= \frac{1}{2} \cdot 6\cdot 10\cdot sin110^0=30\cdot sin (180^0-70^0)=\\\\.\ \ \ \ \ \ =30\cdot sin70^0\approx30\cdot 0.9397=28.191\ [u^2]\\\\

2.\ \ \ |\angle \ ABC|=130^0\\\\.\ \ \ \ \ a=|BC|=92\ [u]\ \ \ and\ \ \ c=|AB|=30\ [u]\\\\Area= \frac{1}{2} \cdot |AB|\cdot |BC|\cdot sin(|\angle \ ABC|)\\\\Area= \frac{1}{2} \cdot 30\cdot 92\cdot sin130^0=1380\cdot sin (180^0-50^0)=\\\\.\ \ \ \ \ \ =1380\cdot sin50^0\approx1380\cdot 0.7660=1057.08\ [u^2]\\\\
5 0
3 years ago
Geometry help will give brainliest
MrRissso [65]
Problem 1
=========
Write an equation for line passing through (8,12) that is perpendicular to 
y = (4/3)x - 5
 
Note:
When two lines are perpendicular, the product of their slopes is -1.

The slope of the given line is 4/3.
Therefore the slope of the perpendicular line is -3/4.
Let the equation of the perpendicular line be
y = -(3/4)x + c
Because the line passes through (8,12), therefore
-(3/4)*8 + c = 12
-6 + c = 12
c = 18
The equation is y = -(3/4)x +18

Answer: y= -\frac{3}{4} x + 18

Problem 2
=========
Write an equation for a line passing through (-30,7) that is perpendicular to
y = -3x - 5.

The perpendicular line will have a slope of 1/3. Let its equation be
y = (1/3)x + c
Because the line passes through (-30,7), therefore
(1/3)*(-30) + c = 7
-10 + c = 7
c = 17
The equation is y = (1/3)x + 17

Answer: y =  \frac{1}{3} x + 17

7 0
3 years ago
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