I need to be walked through this please.

Which sequences of transformations will map circle M onto circle N?

sergiy2304 [10]

Answer:

1: Reflect M across the x-axis

2: Dilate about the center by 3/2

Step-by-step explanation:

Given

See attachment for M and N

Required

Which maps M to N

The coordinates of the radius of the circles are:

$M = (5,5)$

$N = (5,-5)$

And the radius of circles are:

$r_M=2$

$r_N=3$

The first transformation from M to M' is:

- Reflect across the x-axis

The rule is:

$(x,y) \to (x,-y)$

$M(5,5) \to M'(5,-5)$

At this point, M' and N now have the same center but different radius.

The second transformation from M' to N is:

- Dilate about the center by dividing the radius of N by the radius of M

i.e.

$k =\frac{r_N}{r_M}$

$k =\frac{3}{2}$

At this point, M has been completely mapped to N.

4 0

3 years ago

1.)

NemiM [27]

Answer:

There are no specified Numbers provided

Step-by-step explanation:

3 0

2 years ago

Hi, I really need help with my homework please help

8090 [49]

Number one is 68 you have to divide on this download photomath its better because all you have to do is take a picture and it shows you how to do it ;)

5 0

3 years ago

Read 2 more answers

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$

so the left side is

$2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n$

Also recall that

$\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}$

so that the right side is

$b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)$

Solve for $c$ .

$2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}$

Now, the numerator increases more slowly than the denominator, since

$\dfrac{d}{dn}(2n(n-1)) = 4n - 2$

$\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2$

and for $n\ge5$ ,

$2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2$

This means we only need to check if the claim is true for any $n\in\{1,2,3,4\}$ .

$n=1$ doesn't work, since that makes $c=0$ .

If $n=2$ , then

$c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0$

If $n=3$ , then

$c = \dfrac{12}{2^3 - 6 - 1} = 12$

If $n=4$ , then

$c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N$

There is only one value for which the claim is true, $c=12$ .

3 0

2 years ago

A swimming pool is in the shape of a rectangular prism. the pool is 4 feet deep. its length is 4 times its width. the volume of

makvit [3.9K]

First you will have to use the equation length*width*height. The just plug in 4 for the height, then since the length is 4 times longer plug in 4x, and the width would just be x. SO your equation would be 4*4x*x=6400 ft^3. Then foil it out to get 16x^2=6400, then square root 6400 which will give you80, then divide by 16 so x=5 feet^3. That is your width but you length is 4 times that amount. So for length just do 4(5)=20 feet^3

7 0

3 years ago