I did the math, which wasn't too hard if you knew how to solve it.

There will be 14 roses, which is worth $28. 14($2)

And finally there will be 17 daisies, which is worth $17. 17($1)

$28+$17=$45

please vote my answer brainliest. thanks!

8 0

3 years ago

Triangle PQR is transformed to triangle P'Q'R'. Triangle PQR has vertices P(8, 0), Q(6, 2), and R(−2, −4). Triangle P'Q'R' has v

almond37 [142]

Answer:

Part A: The scale factor is 2

Part B:

P''(-4, 0)

Q''(-3, 1)

R''(1, -2)

Part C: No

Step-by-step explanation:

Part A:

The coordinates of the triangle PQR are P(8, 0), Q(6, 2), and R(-2, -4)

The coordinates of the triangle P'Q'R' are P'(4, 0), Q'(3, 1), and R'(-1, -2)

Please find attached the plot of triangles PQR and P'Q'R'

The length of the sides are given by the following formula;

Using an online length

For PQ. the length =

For PR. the length =

For RQ. the length =

While for triangle P'Q'R', we have;

For P'Q'. the length =

For P'R'. the length =

For R'Q'. the length =

∴ PQ = 2 × P'Q'

PR = 2 × P'R'

QR = 2 × Q'R'

Therefore, the scale factor = 2

Part B

By reflecting about the y-axis we have;

Coordinates of the preimage (x, y) after reflection across the y=axis becomes (-x, y)

Therefore, the coordinates of the triangle P''Q''R'' are;

P'(4, 0) after reflection = P''(-4, 0)

Q'(3, 1) after reflection about the y-axis is Q''(-3, 1)

R'(-1, -2) after reflection about the y-axis becomes R''(1, -2)

Part C: No

Given that the triangle P''Q''R'' is congruent to triangle P'Q'R', and triangle P'Q'R' is similar to but not congruent to triangle PQR, we have that triangle P''Q''R'' is similar to but not congruent to triangle PQR.

Step-by-step explanation:

3 0

3 years ago

Can somebody explain how trigonometric form polar equations are divided/multiplied?

koban [17]

Answer:

Attachment 1 : Option C

Attachment 2 : Option A

Step-by-step explanation:

( 1 ) Expressing the product of z1 and z2 would be as follows,

$14\left[\cos \left(\frac{\pi \:}{5}\right)+i\sin \left(\frac{\pi \:\:}{5}\right)\right]\cdot \:2\sqrt{2}\left[\cos \left(\frac{3\pi \:}{2}\right)+i\sin \left(\frac{3\pi \:\:}{2}\right)\right]$

Now to solve such problems, you will need to know what cos(π / 5) is, sin(π / 5) etc. If you don't know their exact value, I would recommend you use a calculator,

cos(π / 5) = $\frac{\sqrt{5}+1}{4}$ ,

sin(π / 5) = $\frac{\sqrt{2}\sqrt{5-\sqrt{5}}}{4}$

cos(3π / 2) = 0,

sin(3π / 2) = - 1

Let's substitute those values in our expression,

$14\left[\frac{\sqrt{5}+1}{4}+i\frac{\sqrt{2}\sqrt{5-\sqrt{5}}}{4}\right]\cdot \:2\sqrt{2}\left[0-i\right]$

And now simplify the expression,

$14\sqrt{5-\sqrt{5}}+i\left(-7\sqrt{10}-7\sqrt{2}\right)$

The exact value of $14\sqrt{5-\sqrt{5}}$ = $23.27510\dots$ and $(-7\sqrt{10}-7\sqrt{2}\right))$ = $-32.03543\dots$ Therefore we have the expression $23.27510 - 32.03543i$ , which is close to option c. As you can see they approximated the solution.

( 2 ) Here we will apply the following trivial identities,

cos(π / 3) = $\frac{1}{2}$ ,

sin(π / 3) = $\frac{\sqrt{3}}{2}$ ,

cos(- π / 6) = $\frac{\sqrt{3}}{2}$ ,

sin(- π / 6) = $-\frac{1}{2}$

Substitute into the following expression, representing the quotient of the given values of z1 and z2,

$15\left[cos\left(\frac{\pi \:}{3}\right)+isin\left(\frac{\pi \:\:}{3}\right)\right] \div \:3\sqrt{2}\left[cos\left(\frac{-\pi \:}{6}\right)+isin\left(\frac{-\pi \:\:}{6}\right)\right]$ ⇒