PLS HELP!! What is the value of X

A traingle has vertices at A(-7, 6), B(4, 9), C(-2, -3). What are the coordinates of each vertex if the triangle is translated 4

Alenkinab [10]

Answer:

The coordinates of each vertex are

A'(-3, 0),B'(8,3) and C'(2,-9)

Step-by-step explanation:

we know that

The triangle is translated 4 units right and 6 units down

That means

The rule for the translation is equal to

(x,y) -------> (x+4,y-6)

Apply the rule of the translation to the coordinates of vertices ABC to obtain the vertices of the image A'B'C'

so

A(-7, 6) -----> A'(-7+4, 6-6)

A(-7, 6) -----> A'(-3, 0)

B(4, 9) ----> B'(4+4,9-6)

B(4, 9) ----> B'(8,3)

C(-2, -3) ----> C'(-2+4,-3-6)

C(-2, -3) ----> C'(2,-9)

3 0

4 years ago

Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

2πr = 60-y

$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0

4 years ago

30 POINTS!

Goshia [24]

Answer:

The correct answare would be 3.54 x 10^22 N

7 0

3 years ago

Who wants 50 point..

Alina [70]

Ayyyyeeeeeeeeeee let’s gooooooooooooo babbbbyyyyyyyyyyyy

7 0

3 years ago

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Find the vertex of the function given below. y = x2 - 2x + 1

posledela

The formula for the Axis of symmetry is -b/2a. Which is -(-2)/2(1) = 2/2 = 1. The axis of symmetry is the X of the vertex, so just plug the 1 into the equation. So the Y= 0, so the vertex is (1 , 0)

6 0

4 years ago

Read 2 more answers