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3 years ago

Finish the rule for a transformation that translates 2 units up and 3 units left. (x,y) →

Mathematics

1 answer:

kykrilka [37]3 years ago

5 0

Answer:

(x - 3, y + 2)

Step-by-step explanation:

When we are translating up/down, we move y.

When we are translating left/right, we move x.

Therefore, if we are moving up 2 and left 3:

Up = +2 to y

Left = -3 to x

(x - 3, y + 2)

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When trying to find the probability of rolling an even number on a number cube with numbers 1 through 6, how many desired outcom

shepuryov [24]

Answer:

Step-by-step explanation:

3 even numbers 2,4,6

or 1/2

or 50%

7 0

3 years ago

What is 7.42 as a fraction in simplest form?

inn [45]

7.42 as a fraction is 7420/1000. Now you need to simplify it. The easiest way to do this is to continuously divide the top and bottom by 2 as long as it is even. At this point it should be small enough to figure out the other factors, and finish simplifying it. This only simplifies to 371/50. While 371 is divisible by 7 and 53, 50 is not divisible by either of them, so the final solution is 371/30

Therefore 371/50 is 7.42 as a fraction.

8 0

3 years ago

Read 2 more answers

The rectangle shown is enlarged such that each side is multiplied by the value of the width, 2x. Which expression represents the

kykrilka [37]

Answer:

<h2> The answer is </h2><h2> D. 8x^2+4xy</h2>

Step-by-step explanation:

We know that the expression for the perimeter is given as

P=2l+2B

now given that the value of the width is =2x

length= 4x

and the breadth=y

P=4x+2y

let us multiply both the length and the width with 2x we have

P=2x(4x+2y)

P=8x^2+4xy

the answer is

D. 8x^2+4xy

4 0

3 years ago

If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.

Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have $S_1=3\times2^{1-1}+2(-1)^1=3-2=1$ , which agrees with the initial value S₁ = 1.

• Assume the closed form is correct for all n up to n = k. In particular, we assume

$S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}$

and

$S_k=3\times2^{k-1}+2(-1)^k$

We want to then use this assumption to show the closed form is correct for n = k + 1, or

$S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}$

From the given recurrence, we know

$S_{k+1}=S_k+2S_{k-1}$

so that

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)$

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}$

$S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)$

$S_{k+1}=3\times2^k-2(-1)^k$

$S_{k+1}=3\times2^k+2(-1)(-1)^k$

$\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}$

which is what we needed. QED

6 0

3 years ago

Find (r-s) (t-s) + (s-r) (s-t) for all numbers r, s, and t. (a) 0 (b) 2 (c) 2rt (d) 2(s-r) (t-s) (e) 2(r-s) (t-s)

lyudmila [28]

Notice that the 2 expressions have 2 common terms.

(r-s) is just (s-r) times (-1)

similarly

(t-s) is just (s-t) times (-1)

this means that :

(r-s) (t-s) + (s-r) (s-t)=-(s-r)[-(s-t)]+(s-r) (s-t)

the 2 minuses in the first multiplication cancel each other so we have:

-(s-r)[-(s-t)]+(s-r) (s-t)=(s-r) (s-t)+(s-r) (s-t)=2(s-r) (s-t)

Answer:

d)2(s-r) (t-s)

3 0

3 years ago

Read 2 more answers