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2 years ago

Which rule describes a translation that is 6 units to the left and 4 units up?

Mathematics

2 answers:

Korvikt [17]2 years ago

6 0

Unit 4 Lesson 9 Unit Test "Transformations"

1) A. I and III only

2) C.

3) B. (x,y) arrow (x-6, y-5)

4) D. (4,-5)

5) C. P'(4,1), Q'(2,-8), and R'(-8,-1)

6) D.

7) A. ON

8) A. 240 Degrees

9) D. 6

10) C. S

11) C. Circle

12) D. No symmetry

13) A. (5,9)

14) D is flipped when reflected over l and is regular when reflected over m

15) Translation; does not rotate just slides down and is not altered

16)

17) A(-3,1)=(-12,4), B(4,-3)=(16,-12) C(2,3)=(8,12) D(-1,4)=(-4,16) Since you multiply each by the scale factor of 4.

All multi choice are 100% correct I just took the test.

Dimas [21]2 years ago

5 0

<span> x represents displacements which are positive to the right of the origin, while y represents displacements which are positive when vertically upward. </span>

<span>Then the required displacement has x decreasing by 6 units, and y increasing by 5 units. (x – 6, y + 5) is then the correct answer</span>

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K(n) =4n + 1; find k(-7)

yarga [219]

Answer:

k(-7) = -27

Step-by-step explanation:

K(n) = 4x + 1

find k(-7)

so we have 4(-7 ) +1

-28 +1

Therefore, k(-7) = -27

3 0

2 years ago

If 8x−5y=−3 is a true equation, what would be the value of 8x−5y+7?

Vesna [10]

Answer:

Value of 8x -5y + 7 is 4.

Step-by-step explanation:

Given that:

8x-5y = -3 Eqn 1

It is a true equation.

Now we have to find the value of the given expression

8x -5y + 7

Putting value of 8x-5y from Eqn 1 in the given expression

8x - 5y + 7

-3 + 7

Therefore,

8x - 5y + 7 = 4

Hence,

Value of 8x -5y + 7 is 4.

7 0

2 years ago

Which is the equation of a hyperbola with directrices at x = ±2 and foci at (5, 0) and (−5, 0)? y squared over 40 minus x square

N76 [4]

The equation of the hyperbola with directrices at x = ±2 and foci at (5, 0) and (−5, 0) is $\frac{x^2}{10} + \frac{y^2}{15} = 1$

<h3>How to determine the equation of the hyperbola?</h3>

The given parameters are:

Directrices at x = ±2
Foci at (5, 0) and (−5, 0)

The foci of a hyperbola are represented as:

Foci = (k ± c, h)

The center is:

Center = (h,k)

And the directrix is:

Directrix, x = h ± a²/c

By comparison, we have:

k ± c = ±5

h = 0

h ± a²/c = ±2

Substitute h = 0 in h ± a²/c = ±2

0 ± a²/c = ±2

This gives

a²/c = 2

Multiply both sides by c

a² = 2c

k ± c = ±5 means that:

k ± c = 0 ± 5

By comparison, we have:

k = 0 and c = 5

Substitute c = 5 in a² = 2c

a² = 2 * 5

a² = 10

Next, we calculate b using:

b² = c² - a²

This gives

b² = 5² - 10

Evaluate

b² = 15

The hyperbola is represented as:

$\frac{(x - k)^2}{a^2} + \frac{(y - h)^2}{b^2} = 1$

So, we have:

$\frac{(x - 0)^2}{10} + \frac{(y - 0)^2}{15} = 1$

Evaluate

$\frac{x^2}{10} + \frac{y^2}{15} = 1$

Hence, the equation of the hyperbola is $\frac{x^2}{10} + \frac{y^2}{15} = 1$