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2 years ago

Write the rule for the translation below. Draw to show your thinking

Mathematics

1 answer:

frez [133]2 years ago

3 0

Answer:

a translation shifts each point the same distance horizontally and the same distance vertically.

Step-by-step explanation:

You might be interested in

(p^2-7pq-q^2)+(-3p^2-2pq+7q^2)<br><br>

Alex Ar [27]

Answer:

-2p² - 9pq + 6q²

Step-by-step explanation:

Add the like terms

p² + - 3p² = -2p²

-7pq + - 2pq = -9pq

-q² + 7q² = 6q²

6 0

3 years ago

Add: (15a + 3b) + (a + 3b – 4)

Nookie1986 [14]

16a + 15b is the answer

6 0

3 years ago

Evaluate the equation. -4x = 14 A.10 B. 7/2 C. -7/2 D. -10

Alex_Xolod [135]

Answer:

c) -7/2

Step-by-step explanation:

14/-4= -3.5

4 0

3 years ago

Read 2 more answers

22. f(x) is stretched horizontally by a factor of 2 and reflected across the x-axis. Which choice shows the correct representati

goblinko [34]

Answer:

A. -f(1/2 x)

Step-by-step explanation:

Reflextion about the x-axis is

f(x) -> -f(x)

and horizontal dilation is

f(x) -> f(-x/b) where b is the factor of dilation.

so the proper answwer is

A. -f(1/2 x)

6 0

3 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.

3 0

3 years ago