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iren [92.7K]
3 years ago
13

The points P(−3, 6) and Q(1, 4) are two vertices of right triangle PQR . The hypotenuse of the triangle is PQ⎯⎯⎯⎯⎯ .

Mathematics
2 answers:
Anna007 [38]3 years ago
6 0

Answer: Coordinates of point R is (1,6)

Explanation:

Since we have given that

The points P(-3,6) and Q(1,4) are two vertices of right triangle PQR.

The hypotenuse of the triangle is PQ.

Construct  a rectangle PQRR' with diagonal PQ ,

Now, we'll determine the points of R and R',

With the help of P(-3,6) and Q(1,4)

R will be (1,6) as x-coordinate of R and Q will be same and y-coordinate of R will be while determining the coordinates of R.

Similarly,

With the help of P(-3,6) and Q(1,4)

R' will be (-3,4) as x-coordinate of P and R' will be same and y-coordinate of Q and R' will be same while determining the coordinates of R'.

But as we given that point R lies in quadrant 1 so, we will take the value of R which (1,6).

Hence, coordinates of point R is (1,6)

NARA [144]3 years ago
3 0
The answer is R (1,6.)
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Answer:

  substitute that value for x in the polynomial and see if it evaluates to zero

Step-by-step explanation:

A "zero" of a polynomial is a value of the polynomial's variable that make the expression become zero when it is evaluated. As an almost trivial example, consider the polynomial x-3. The value x = 3 is a zero because substituting that value for x makes the expression evaluate as zero.

  3 -3 = 0

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6 0
3 years ago
Y=-3|x+1|+1 ,what is the slope and vertex?
hodyreva [135]

Answer:

Alright well Graph the absolute value using the vertex and a few selected points

X: -3, -2, -1, 0, 1

Y: -5, -2, 1, -2, -5 . Hope this helps :)

Step-by-step explanation:


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3 years ago
What is the intermediate step in the form square for the following equation? (x + a) ^ 2 = b as a result of completing the squar
Ivahew [28]

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ndendnnnndbdndn nnnnd

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2 years ago
What is the equation of the line that passes through the points (4/5,1/5) and (1/2,3/2)?​
Setler [38]

Answer:

y-\frac{3}{2}=\frac{-13}{3}(x-\frac{1}{2}) point-slope form

13x+3y=11 (standard form)

Let me know if you prefer another form.

Step-by-step explanation:

The slope of a line can be found using \frac{y_2-y_1}{x_2-x_1} provided you are given two points on the line.

We are.

Now you can use that formula.  But I really love to just line up the points vertically then subtract them vertically then put 2nd difference over 1st difference.

 (4/5  ,  1/5)

-( 1/2  ,  3/2)

-----------------

3/10          -13/10

2nd/1st = \frac{\frac{-13}{10}}{\frac{3}{10}}=\frac{-13}{3} is our slope.

So the following is point-slope form for a linear equaiton:

y-y_1=m(x-x_1) \text{ where } m \text{ is slope and } (x_1,y_1) \text{ is a point on the line }    

Plug in a point (x_1,y_1)=(\frac{1}{2},\frac{3}{2}) \text{ and } m=\frac{-13}{3}.

This gives:

y-\frac{3}{2}=\frac{-13}{3}(x-\frac{1}{2})

I'm going to distribute:

y-\frac{3}{2}=\frac{-13}{3}x-\frac{-13}{6}

Now I don't like these fractions so I'm going to multiply both sides by the least common multiply of 2,3, and 6 which is 6:

6y-9=-26x+13

Add 26x on both sides:

26x+6y-9=13

Add 9 on both sides:

26x+6y=22 This is actually standard form of a line.

It can be simplified though.

Divide both sides by 2:

13x+3y=11 (standard form)

7 0
3 years ago
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Katen [24]

Answer:

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Step-by-step explanation:

if you add them

7 0
2 years ago
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