PLS ANSWER!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

the vertices of PQR are P(-5,1), Q(-4,6), and R(-2,3), graph P"Q"R" after a composition of the transformations in the order they

horsena [70]

SOLUTION

We are told to translate; (x, y) to (x -8, y). This means we have to add - 8 to each value of x in P(-5,1), Q(-4,6), and R(-2,3).

In P(-5,1), x = -5 and y = 1

In Q(-4,6), x = -4 and y = 6 and

In R(-2,3), x = -2 and y = 3

$\begin{gathered} P(-5,\text{ 1) translates to (-5 -8, 1) }\rightarrow P^i(-13,\text{ 1)} \\ Q(-4,\text{ 6) translates to (-4 -8, 6) }\rightarrow Q^i(-12,\text{ 6)} \\ R(-2,\text{ 3) translates to (-2 -8, 3) }\rightarrow R^i(-10,\text{ 3)} \end{gathered}$

For the dilation centered at the origin k =2, simply multiply the value of k, which is 2 into the translations.

$\begin{gathered} At\text{ K = 2, }P^i(-13,\text{ 1) }\rightarrow\text{ }2(-13,\text{ 1) }\rightarrow\text{ }P^{ii}(-26,\text{ 2)} \\ Q^i(-12,\text{ 6) }\rightarrow\text{ }2(-12,\text{ 6) }\rightarrow\text{ Q}^{ii}(-24,\text{ 12)} \\ R^i(-10,\text{ 3) }\rightarrow\text{ }2(-10,\text{ 3) }\rightarrow\text{ R}^{ii}(-20,\text{ 6)} \end{gathered}$

5 0

2 years ago

Help ASSAP with this question

mote1985 [20]

(4,4)(-6,4)...notice how ur y coordinates on both of ur points is the same...that means u have a horizontal line with a slope of 0....the equation would be
y = 4.....but in point slope form...not 100% sure

y - y1 = m(x - x1)
slope(m) = 0
using (4,4)...x1 = 4 and y1 = 4
now sub
y - 4 = 0(x - 4) <==

y - y1 = m(x - x1)
slope(m) = 0
using (-6,4)...x1 = -6 and y1 = 4
sub
y - 4 = 0(x - (-6) =
y - 4 = 0(x + 6) <==

6 0

4 years ago

What is the answer to solving <br> for y -8=y-9

Makovka662 [10]

Answer:

no solution

Step-by-step explanation:

y-8=y-9

y=y-1

error

6 0

3 years ago

a 16 inch wire is to be cut. one piece is to be bent into the shape of a square, whereas the other piece is to be bent into the

Lyrx [107]

1. Let x be a length of square side, then the square perimeter is 4x (you need to cut 4x from 16 in. to make a square with side x in.). Then 16-4x is remained length of wire and you have to make form this piece a rectangle with sides one of which is twice bigger than another. If y is length of the smaller side, then 2y is length of the bigger side and rectangle perimeter is y+2y+y+2y=6y. You have 16-4x in. of wire left, so 6y=16-4x and $y= \frac{16-4x}{6}$ .

2.
$A_{square}=x^2 \\ A_{rectangle}=y\cdot 2y=\frac{16-4x}{6}\cdot2\cdot\frac{16-4x}{6}= \frac{(16-4x)^2}{18} \\ A(x)=A_{square}+A_{rectangle}=x^2+\frac{(16-4x)^2}{18}$ .

3. Find the derivative of the function A(x):
$A'(x)=2x+ \dfrac{2(16-4x)\cdot(-4)}{18} =2x- \dfrac{4(16-4x )}{9}$ .

4. Solve the equation A'(x)=0:
$2x- \dfrac{4(16-4x )}{9}=0 \\ 18x-64+16x=0 \\ 34x=64 \\ x= \frac{32}{17}$

5. Since $y= \frac{16-4x}{6}$ you have

$y= \dfrac{16-4 \frac{32}{17} }{6} = \dfrac{16\cdot17-4\cdot32}{6\cdot 17} = \dfrac{24}{17}$ .

Answer: <span>the width of the rectangle is $\frac{24}{17}$ </span>

6 0

3 years ago

Technetium-99m, a radioisotope used to image the skeleton and the heart muscle, has a half-life of about 6 hours. Find the decay

sashaice [31]

The decay constant is i 0.1155, and there would be 16 mg left after 24 hours.

The relationship between the half-life, T₀.₅, and the decay constant, λ, is given by

T₀.₅ = 0.693/λ.

Solving for λ, we will multiply both sides by λ first:
(T₀.₅)(λ) = 0.693

Since we know the half life is 6 hours, this gives us:
6λ = 0.693

Dividing by 6, we have
λ = 0.693/6 = 0.1155.

The decay constant will be k in our decay formula, and N₀, the original amount of substance, is 250:
N(24) = 250e^(-0.1155*24) = 15.6 ≈ 16

7 0

3 years ago