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3 years ago

7

circle on has a center of (-4, -7) and a radius of 12cm. circle 2 has a center of (3, 4) and a radius of 15cm. what transformati

ons can be applied to circle one to prove that the circles are similar?

1 answer:

AlekseyPX3 years ago

3 0

-4, -21, 4. hope this helps.

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Please help asap - will give 20 points

Ludmilka [50]

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3 more weeks. this is the answer. just took the assignment

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4 years ago

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Let x be a random variable representing percentage change in neighborhood population in the past few years, and let y be a rando

Lisa [10]

Answer:

$m=\frac{2506.833}{604.833}=4.1447$

Nowe we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{77}{6}=12.833$

$\bar y= \frac{\sum y_i}{n}=\frac{587}{6}=97.833$

And we can find the intercept using this:

$b=\bar y -m \bar x=97.833-(4.1447*12.833)=44.644$

So the line would be given by:

$y=4.1447 x +44.644$

Step-by-step explanation:

$m=\frac{S_{xy}}{S_{xx}}$

Σx = 77, Σy = 587, Σx2 = 1593, Σy2 = 70,533, and Σxy = 10040.

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=1593-\frac{77^2}{6}=604.833$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=10040-\frac{77*587}{6}=2506.833$

And the slope would be:

$m=\frac{2506.833}{604.833}=4.1447$

Nowe we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{77}{6}=12.833$

$\bar y= \frac{\sum y_i}{n}=\frac{587}{6}=97.833$

And we can find the intercept using this:

$b=\bar y -m \bar x=97.833-(4.1447*12.833)=44.644$

So the line would be given by:

$y=4.1447 x +44.644$

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3 years ago

8(1/4)-1+0.5(10)<br> please help meeeeeee

maks197457 [2]

Answer:

6

Step-by-step explanation:

8(1/4) is 2 bc 1/4 of 8 is 2

2-1=.5(10)

.5 works like 1/2 so 1/2 is 10 is 5

then you get 2-1+5 then after that you get 6

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3 years ago

In ΔRST, s = 93 inches, ∠S=123° and ∠T=28°. Find the length of r, to the nearest 10th of an inch.

FrozenT [24]

We have been given that in ΔRST, s = 93 inches, ∠S=123° and ∠T=28°. We are asked to find the length of r to the nearest 10th of an inch.

We will use law of sines to solve for side r.

$\frac{a}{\text{Sin}(a)}=\frac{b}{\text{Sin}(B)}=\frac{c}{\text{Sin}(C)}$ , where a, b and c are corresponding sides to angles A, B and C respectively.

Let us find measure of angle S using angle sum property of triangles.

$\angle R+\angle S+\angle T=180^{\circ}$

$\angle R+123^{\circ}+28^{\circ}=180^{\circ}$

$\angle R+151^{\circ}=180^{\circ}$

$\angle R+151^{\circ}-151^{\circ}=180^{\circ}-151^{\circ}$

$\angle R=29^{\circ}$

$\frac{r}{\text{sin}(R)}=\frac{s}{\text{sin}(S)}$

$\frac{r}{\text{sin}(29^{\circ})}=\frac{93}{\text{sin}(123^{\circ})}$

$\frac{r}{\text{sin}(29^{\circ})}\cdot \text{sin}(29^{\circ})=\frac{93}{\text{sin}(123^{\circ})}\cdot \text{sin}(29^{\circ})$

$r=\frac{93}{0.838670567945}\cdot (0.484809620246)$

$r=110.889786233799179\cdot (0.484809620246)$

$r=53.7604351$

Upon rounding to nearest tenth, we will get:

$r\approx 53.8$

Therefore, the length of r is approximately 53.8 inches.

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