All categories

3 years ago

What coordinates for point c would make pqt simular to abc

Mathematics

1 answer:

Dafna1 [17]3 years ago

5 0

Answer:

A. (7, -5)

Step-by-step explanation:

The given coordinates of the vertices of ΔPQT are;

P(-2, 5), Q(-2, 1) and T(-5, 1)

The length of the side TQ = 3 units

The length of the side PQ = 4 units

Therefore, the length of the side PT = √(3² + 4²) = 5

The coordinates of the line segment AB = A(1, 3), B(1, -5)

Therefore, using a scale factor of 2, where the sides PQ and AB are corresponding sides, we have;

Where BC is the corresponding side to the TQ on ΔPQT, we have;

The length of BC = 2 × The length of TQ = 2 × 3 units = 6 units (distant from B along the x-axis)

Therefore, the possible points for the point C are;

C(1 - 6, -5) or C(1 + 6, -5)

C(-5, -5) or C(7, -5)

Therefore, the correct option is option A. C(7, -5).

You might be interested in

3/4 divides by 1/7 =

anyanavicka [17]

3/4 divides by 1/7 =

copy dot flip

3/4 * 7/1

21/4

4 goes into 21 5 times with 1 left over

5 1/4

4 0

3 years ago

Read 2 more answers

Hey someone who knows this want to give me the answer thanks

Gnesinka [82]

C is the answer. Hope this helps!

4 0

3 years ago

Read 2 more answers

My last question plz help me

LenKa [72]

The answer is c I hope this helps

4 0

3 years ago

Read 2 more answers

These points are linear. Find the slope. x0 |1 |2|3|4|5 y|0|2|4|6|8|10 slope = [?]

jolli1 [7]

Answer:

Step-by-step explanation:

x₁ = 0 ; y₁=0 &

x₂ = 1 ; y₂=2

Slope = $\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$=\dfrac{2-0}{1-0}\\\\\\=\dfrac{2}{1}\\\\\\= 2$

Slope = 2

7 0

3 years ago

Write an equation for a quadratic function that has x intercepts (-3, 0) and (5, 0)

Blizzard [7]

Answer:

One possible equation is $f(x) = (x + 3)\, (x - 5)$ , which is equivalent to $f(x) = x^{2} - 2\, x - 15$ .

Step-by-step explanation:

The factor theorem states that if $x = x_{0}$ (where $x_{0}$ is a constant) is a root of a function, $(x - x_{0})$ would be a factor of that function.

The question states that $(-3,\, 0)$ and $(5,\, 0)$ are $x$ -intercepts of this function. In other words, $x = -3$ and $x = 5$ would both set the value of this quadratic function to $0$ . Thus, $x = -3\!$ and $x = 5\!$ would be two roots of this function.

By the factor theorem, $(x - (-3))$ and $(x - 5)$ would be two factors of this function.

Because the function in this question is quadratic, $(x - (-3))$ and $(x - 5)$ would be the only two factors of this function. In other words, for some constant $a$ ( $a \ne 0$ ):

$f(x) = a\, (x - (-3))\, (x - 5)$ .

Simplify to obtain:

$f(x) = a\, (x + 3)\, (x - 5)$ .

Expand this expression to obtain:

$f(x) = a\, (x^{2} - 2\, x - 15)$ .

(Quadratic functions are polynomials of degree two. If this function has any factor other than $(x - (-3))$ and $(x - 5)$ , expanding the expression would give a polynomial of degree at least three- not quadratic.)

Every non-zero value of $a$ corresponds to a distinct quadratic function with $x$ -intercepts $(-3,\, 0)$ and $(5,\, 0)$ . For example, with $a = 1$ :

$f(x) = (x + 3)\, (x - 5)$ , or equivalently,

$f(x) = x^{2} - 2\, x - 15$ .

6 0

2 years ago