All categories

2 years ago

Simple answers please

Mathematics

1 answer:

charle [14.2K]2 years ago

4 0

Answer:

A'(4, -4)
B'(0, -3)
C'(2, -1)
D'(3, -2)

Step-by-step explanation:

The coordinate transformation for a 270° clockwise rotation is the same as for a 90° counterclockwise rotation:

(x, y) ⇒ (-y, x)

The rotated points are ...

A(-4, -4) ⇒ A'(4, -4)

B(-3, 0) ⇒ B'(0, -3)

C(-1, -2) ⇒ C'(2, -1)

D(-2, -3) ⇒ D'(3, -2)

_____

Additional comment

To derive and/or remember these transformations, it might be useful to consider where a point came from when it ends up on the x- or y-axis.

A point must have come from the -y axis if rotating it 270° CW makes it end up on the +x-axis. A point must have come from the x-axis if rotating it 270° makes it end up on the +y axis. That is why we write ...

(x, y) ⇒ (-y, x) . . . . . . the new x came from -y; the new y came from x

You might be interested in

G divided by 93 is 264

Tanya [424]

See picture for solution steps and answer.

4 0

3 years ago

Read 2 more answers

In a certain Algebra 2 class of 25 students, 6 of them play basketball and 12 of them play baseball. There are 11 students who p

bixtya [17]

Answer:

4/25

Step-by-step explanation:

7 0

3 years ago

Determine whether the series is convergent or divergent. 1 4 + 3 16 + 1 64 + 3 256 + 1 1024 + 3 4096 +

Alla [95]

This series converges.

$S=\dfrac14+\dfrac3{16}+\dfrac1{64}+\dfrac3{256}+\dfrac1{1024}+\cdots$

$S=\left(\dfrac34+\dfrac3{16}+\dfrac3{64}+\dfrac3{256}+\dfrac3{1024}+\cdots\right)-\left(\dfrac12+\dfrac1{32}+\dfrac1{512}+\cdots\right)$

$S=\displaystyle\sum_{n=1}^\infty\frac3{4^n}-\sum_{n=0}^\infty\frac1{2^{4n+1}}$

Both component series are geometric with ratios less than 1, so they both converge.

$\displaystyle\sum_{n=1}^\infty\frac3{4^n}=1$

$\displaystyle\sum_{n=0}^\infty\frac1{2^{4n+1}}=\frac12\sum_{n=0}^\infty\frac1{16^n}=\frac8{15}$

So we have

$S=\dfrac7{15}$

8 0

3 years ago

How do u solve the formula for the area of a triangle

kifflom [539]

Answer:

A = bh/2

Solved for b:

b = 2A/h

Solved for h:

h = 2A/b

3 0

3 years ago

Evaluate (6 + 2i) / (3 –i ). 5/4 + 6/4i 8/5 + 6/5i 20/8 + 12/8i 16/9 + 12/9i

Yuki888 [10]

We are given with a complex no. and need to simplify it , so let's start !!!

Let's assume that :

${:\implies \quad z=\sf \dfrac{6+2\iota}{3-\iota}}$

Now , Rationalizing the denominator or in other words multiplying and dividing ${z}$ by the conjugate of the denominator

${:\implies \quad z=\sf \dfrac{6+2\iota}{3-\iota}\times \dfrac{3+\iota}{3+\iota}}$

${:\implies \quad z=\sf \dfrac{(6+2\iota)(3+\iota)}{(3-\iota)(3+\iota)}}$

${:\implies \quad z=\sf \dfrac{6(3+\iota)+2\iota (3+\iota)}{(3)^{2}-(\iota)^{2}}\quad \qquad \{\because (a-b)(a+b)=a^{2}-b^{2}\}}$

${:\implies \quad z=\sf \dfrac{18+6\iota +6\iota +2(\iota)^{2}}{9-(-1)}\quad \qquad \{\because (\iota)^{2}=-1\}}$

${:\implies \quad z=\sf \dfrac{18+12\iota -2}{10}}$

${:\implies \quad z=\sf \dfrac{16+12\iota}{10}}$

${:\implies \quad z=\sf \dfrac{16}{10}+\dfrac{12\iota}{10}}$

${:\implies \quad {\bf \therefore}\quad \underline{\underline{z=\bf \dfrac{8}{5}+\dfrac{6}{5}\iota}}}$

Hence , Option B) (8/5) + (6/5)i is correct :D

7 0

2 years ago