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FinnZ [79.3K]
4 years ago
13

If ABC ∼ RST, then STR ∼ BCA. True False

Mathematics
2 answers:
diamong [38]4 years ago
8 0

I think that it is false


Otrada [13]4 years ago
7 0

Answer:

The given statement "STR ∼ BCA" is true.

Step-by-step explanation:

It is given that

\triangle ABC\sim \triangle RST

The corresponding angles of similar triangles are same and the corresponding sides are proportional.

\angle A=\angle R

\angle B=\angle S

\angle C=\angle T

The we can say that

\triangle STR\sim \triangle BCA

Therefore, the given statement "STR ∼ BCA" is true.

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A survey of physicians in 1979 found that some doctors give a placebo to patients who complain of pain for which the physician c
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Can you identify a parallel or perpendicular equation and type the correct code?
AlladinOne [14]

Answer:

The equation of the line that is parallel to  is  and the equation of the line that is perpendicular to  is .

Let be a line whose equation is:

(1)

Whose explicit form is:

(2)

Where:

- Independent variable.

- Dependent variable.

The slope and x-intercept of the line are  and , respectively.

There are two facts:

A line is parallel to other line when the former has the same slope of the latter.

A line is perpendicular to other line when the former has a slope described the following form (), where  is the slope of the former.

Then, the equation of the line that is parallel to  is  and the equation of the line that is perpendicular to  is .

To learn more on lines, we kindly invite to check this verified question: brainly.com/question/2696693

Hope This Helps.

Please Give Brainliest

3 0
3 years ago
Answer correct you get brainliest answer​
zloy xaker [14]

Answer:

y =  {x}^{2}  - 2

Or if you want with the value of h too.

y =  {(x - 0)}^{2}  - 2

Step-by-step explanation:

y = a {(x - h)}^{2}  + k

Find the value of h and k by using the formula.

h =  -  \frac{b}{2a}  \\ k =  \frac{4ac -  {b}^{2} }{4a}

From y = x²-2

a = 1 \\ b = 0 \\ c =  - 2

Substitute these values in the formula.

h =  -  \frac{0}{2(1)}  \\ h = 0

Therefore, h = 0.

k =  \frac{4(1)( - 2) -  {0}^{2} }{4(1)}  \\ k =  \frac{ - 8}{4}  \\ k =  - 2

Therefore, k = - 2.

From the vertex form, the vertex is at (h, k) = (0,-2). Substitute h = 0, a = 1 and k = -2 in the equation.

y = a {(x - h)}^{2}  + k \\ y = 1 {(x - 0)}^{2}  - 2 \\ y =  {(x)}^{2}  - 2 \\ y =  {x}^{2}  - 2

These type of equation where b = 0 can also be both standard and vertex form.

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