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2 years ago

5

Is there a series of rigid transformations that could map A

2 answers:

topjm [15]2 years ago

7 0

Answer:

your answer is C. Yes, Triangle KLP can be reflected across the line containing KP and then translated so that P is mapped to M on edg2020

MaRussiya [10]2 years ago

3 0

Answer:

Yes, A KLP can be reflected across the line containing KP and then translated so that Pis mapped to M.

Step-by-step explanation:

The figure shows two congruent by HA theorem (they have congruent hypotenuses and a pair of congruent angles adjacent to the hypotenuses) triangles KLP and QNM.

A rigid transformation is a transformation which preserves lengths. Reflection, rotation and translation are rigit transformations.

If you reflect triangle KLP across the leg KP and translate it up so that point P coincides with point M , then the image of triangle KLP after these transformations will be triangle QNM.

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2 years ago

The parent function f(x) = x3 is represented by graph A. Graph A is transformed to get graph B and graph C. Write the functions

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2 years ago

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Scott likes to run long distances. He can run 20 \text{ km}20 km20, start text, space, k, m, end text in 858585 minutes. He want

aleksandrvk [35]

Answer:

<em>Scott will take 221 minutes to run 52 km</em>

Step-by-step explanation:

<u>Speed</u>

The speed of an object can be calculated with the formula:

$\displaystyle v=\frac{d}{t}$

Where d is the distance traveled and t is the time taken.

Scott can run d=20 km in 85 minutes. Thus, his speed is:

$\displaystyle v=\frac{20}{85}\ km/min$

Now he wants to know how many minutes it will take him to run d=52 km. Solving the formula for t:

$\displaystyle t=\frac{d}{v}$

Since the speed has been already determined:

$\displaystyle t=\frac{52}{\frac{20}{85}}$

Multiplying by the reciprocal of the denominator:

$\displaystyle t=52*\frac{85}{20}$

t = 221 min

Scott will take 221 minutes to run 52 km

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2 years ago

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In the derivation of the quadratic formula by completing the square, the equation (x+b over 2a)^2=-4ac+b^2 over 4a^2 is created

vredina [299]

Answer:

The result of applying the square root property of equality to this equation is $x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Step-by-step explanation:

Consider the provided equation.

$\left(x+\dfrac{b}{2a}\right)^2=\dfrac{-4ac+b^2}{4a^2}$

As the above equation is formed by perfect square trinomial so simply applying the square root property as shown:

$\sqrt{(x+\dfrac{b}{2a})^2}=\pm \dfrac{\sqrt{-4ac+b^2}}{\sqrt{4a^2}}\\x+\dfrac{b}{2a}=\pm \dfrac{\sqrt{b^2-4ac}}{2a}$

Isolate the variable x.

$x=-\dfrac{b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a}\\x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

Hence, the result of applying the square root property of equality to this equation is $x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

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1 year ago

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