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SCORPION-xisa [38]
2 years ago
15

The image of the point (2, 1) under a translation is (5, -3). Find

Mathematics
1 answer:
Yanka [14]2 years ago
4 0

Answer:

The coordinates of the image of the point (6,6) under the same  translation is: (9, 2)

Hence, option C is correct.

Step-by-step explanation:

The image of the point (2, 1) under a translation is (5, -3).

It means when we horizontally move 3 units to the RIGHT i.e. adding 3 units to the x-coordinate and vertically move 4 units DOWN i.e. subtracting 4 units from the y-coordinate of the original point (2, 1), we get the coordinates of the image (5, -3).

Thus,

The rule of translation can be formulated such as:

(x, y) → (x + 3, y - 4)

(2, 1) → (2 + 3, 1 - 4) → (5, -3)

Thus,

Under the same rule of translation, we can determine the coordinates of the image of the point (6,6):

(x, y) → (x + 3, y - 4)

(6, 6) → (6 + 3, 6 - 4) → (9, 2)

Therefore, the coordinates of the image of the point (6,6) under the same  translation is: (9, 2)

Hence, option C is correct.

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What would the equation be ?
Vikentia [17]

Answer:

y = 4/5x-18/5

Step-by-step explanation:

First we find the slope from the given points (-3,-6) (2,-2)

m = (y2-y1)/(x2-x1)

m = (-2 - -6)/(2 - -3)

   (-2+6)/( 2+3)

  4/5

The slope intercept form of an equation is

y = mx +b  where m is the slope and b is the y intercept

y = 4/5 x +b

Substitute a point into the equation to find b

-6= 4/5(-3) +b

-6= -12/5 +b

Add 12/5 to each side

-6 +12/5 = -12/5 +12/5 +b

Get a common denominator

-30/5 + 12/5 = b

-18/5 =b

The equation is

y = 4/5x-18/5

4 0
3 years ago
A line tangent to the curve f(x)=1/(2^2x) at the point (a, f(a)) has a slope of -1. What is the x-intercept of this tangent?
kirza4 [7]

Answer:

x-intercept = 0.956

Step-by-step explanation:

You have the function f(x) given by:

f(x)=\frac{1}{2^{2x}}   (1)

Furthermore you have that at the point (a,f(a)) the tangent line to that point has a slope of -1.

You first derivative the function f(x):

\frac{df}{dx}=\frac{d}{dx}[\frac{1}{2^{2x}}]  (2)

To solve this derivative you use the following derivative formula:

\frac{d}{dx}b^u=b^ulnb\frac{du}{dx}

For the derivative in (2) you have that b=2 and u=2x. You use the last expression in (2) and you obtain:

\frac{d}{dx}[2^{-2x}]=2^{-2x}(ln2)(-2)

You equal the last result to the value of the slope of the tangent line, because the derivative of a function is also its slope.

-2(ln2)2^{-2x}=-1

Next, from the last equation you can calculate the value of "a", by doing x=a. Furhtermore, by applying properties of logarithms you obtain:

-2(ln2)2^{-2a}=-1 \\\\2^{2a}=2(ln2)=1.386\\\\log_22^{2a}=log_2(1.386)\\\\2a=\frac{log(1.386)}{log(2)}\\\\a=0.235

With this value you calculate f(a):

f(a)=\frac{1}{2^{2(0.235)}}=0.721

Next, you use the general equation of line:

y-y_o=m(x-x_o)

for xo = a = 0.235 and yo = f(a) = 0.721:

y-0.721=(-1)(x-0.235)\\\\y=-x+0.956

The last is the equation of the tangent line at the point (a,f(a)).

Finally, to find the x-intercept you equal the function y to zero and calculate x:

0=-x+0.956\\\\x=0.956

hence, the x-intercept of the tangent line is 0.956

5 0
2 years ago
Paul surveyed students in a high school about their favorite sport of the 1, 200 students in the school, 67 tenth-grade students
Leno4ka [110]

Answer:

2 Paul's Survey sample only included tenth graders

Step-by-step explanation:

The reson why this is your answer is simple

Because this is true, he ONLY surveyed tenth-graders. He should have found for all of the students, including the other 1,100 students that didn't get surveyed.

Thanks!

5 0
2 years ago
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