Write an expression for "the difference of 10 and <br> x.

(02.02 MC) Triangle ABC is shown. A is at negative 2, 1. B is at negative 1, 4. C is at negative 4, 5. If triangle ABC is reflec

Bezzdna [24]

Given:

The vertices of a triangle ABC are A(-2,1), B(-1,4) and C(-4,5).

Triangle ABC is reflected over the x‐axis, reflected over the y‐axis, and rotated 180 degrees.

To find:

The point B'.

Solution:

If a figure reflected over x-axis, then

$(x,y)\to (x,-y)$

$B(-1,4)\to B_1(-1,-4)$

If a figure reflected over y-axis, then

$(x,y)\to (-x,y)$

$B_1(-1,-4)\to B_2(-(-1),-4)$

$B_1(-1,-4)\to B_2(1,-4)$

If a figure rotated 180 degrees about the origin, then

$(x,y)\to (-x,-y)$

$B_2(1,-4)\to B'(-(1),-(-4))$

$B_2(1,-4)\to B'(-1,4)$

So, the coordinate of point B' are (-1,4).

Therefore, the correct option is A.

6 0

3 years ago

The equation of function h is h(x) = 1/2(x - 2)^2. The table shows some of the values of function m.

PtichkaEL [24]

Answer:duuno i wanted to help

Step-by-step explanation:

thnx

3 0

3 years ago

Read 2 more answers

What is the ? in this pattern <br><br><br> 3, -6, 12, 4, 20, ?

SCORPION-xisa [38]

Is skipped a number and that's not good

5 0

3 years ago

GIVING BRAINLIST HELP ASAP PLSS

puteri [66]

Answer:

5: fist one

6: fourth one

7: thrid one

6 0

3 years ago

I need some help with math

Irina18 [472]

Answer:

$(y-7)^2+(x-2)^2=16$

and

$(x+2)^2+(y-15)^2 = 9$

Step-by-step explanation:

The standard equation of a circle is $(x-h)^2+(y-k)^2=r^2$ where the coordinate (h,k) is the center of the circle.

Second Problem:

We can start with the second problem which uses this info very easily.
(h,k) in this problem is (-2,15) simply plug these into the equation. $(x--2)^2+(y-15)^2=r^2$ .
We can also add the radius 3 and square it so it becomes 9. The equation.
This simplifies to $(x+2)^2+(y-15)^2 = 9$ .

First Problem:

The first problem takes a different approach it is not in standard form. But we can convert it to standard form by completing the square.
$y^2-14y+x^2-4x+37=0$ first subtract 37 from both sides so the equation is now $y^2-14y+x^2-4x=-37$ .
$y^2-14y+x^2-4x+37=0$ by adding $(-\frac{b}{2a} )^2$ to both the x and y portions of this equation you can complete the squares. $(-\frac{b}{2a})^2=(-\frac{-14}{2(1)})^2$ and $(-\frac{-4}{2(1)})^2$ which equals 49 and 4.
Add 49 and 4 to both sides and the equation is now: $y^2-14y+49+x^2-4x+4=-37+49+4$ You can simplify the y and x portions of the equations into the perfect squares or factored form $(y-7)^2$ and $(x-2)^2$ .
Finally put the whole thing together. $(y-7)^2+(x-2)^2=16$ .

I hope this helps!

5 0

3 years ago

Write an expression for "the difference of 10 and x.