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4 years ago

Triangle XYZ is rotated 90° clockwise about the origin.

Mathematics

1 answer:

Vlad [161]4 years ago

4 0

Answer:

(1,3)

Step-by-step explanation:

Rotating clockwise means that you're rotating by a negative angle measure. The rule for rotating -90 degrees is (x,y)->(y,-x). X is (-3,1) so X' should be (1,3) since you negate -3 to get 3 and switch the x and y coordinates.

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Salsk061 [2.6K]

Answer:

18 is divisible by both 2 and 9, you can multiply 9 by 2 to get 18, and 18 is a multiple of both of those numbers

Step-by-step explanation:

5 0

3 years ago

If a certain number of Grade IV students share 28 or 39 skill books in Mathematics,there are remainders of 4 and 3 books. What i

nadya68 [22]

Answer:

12 students

Step-by-step explanation:

Below numbers are divisible by the number of students:

28 - 4 = 24
39 - 3 = 36

Lets find GCF of 24 and 36:

24 = 2*2*2*3
36 = 2*2*3*3
GCF(24,36) = 2*2*3 = 12

The largest possible number of students is 12

7 0

3 years ago

What is the slope of the line between (3,-4) and (-2, 1)? O 1 02. O-2 O-1

natka813 [3]

Answer:

-1……..........................

4 0

3 years ago

Read 2 more answers

100 points and brainliest for correct answer!

MatroZZZ [7]

Answer B.

∑fx=1637,∑fx
2
=127663,∑f=21
x
ˉ
=Mean=
∑t
∑fx

=77.95
σ
2
=
∑f
∑fx
2

−(
x
ˉ
)
2
=
21
127663

−(77.95)
2
=2.987
σ=1.728

4 0

2 years ago

Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

4 0

3 years ago