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3 years ago

Is there a series of rigid transformations that could map A

Mathematics

2 answers:

topjm [15]3 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]3 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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Need help with this one also

andriy [413]

Answer:

scale factor 3 is the answer.

Step-by-step explanation:

Length = 18 in

Breadth = 3 in

height = 10 in

Volume of cuboid = l × b × h

V = 18 × 3 × 10

V = 540 in³

540 in³ is the original volume. The volume needs to be tripled, so multiply the original volume with 3

V = 540 × 3 = 1620 in³

∴ 1620 in³ will be the new volume if it's tripled and the scale factor is 3.

4 0

2 years ago

F(x) is a quadratic function with x-intercepts at (−1, 0) and (−3, 0). If the range of f(x) is [−4, ∞) and g(x) = 2x2 + 8x + 6,

svetlana [45]

Answer:

B) Both functions have the same domain.

C) Both functions have the same x-intercepts.

D) Both functions are decreasing on the interval (−∞, −2).

Step-by-step explanation:

step 1

Find the vertex of f(x)

we know that

The x-coordinate of the vertex is the midpoint of the roots

we have

x=-1 and x=-3

the midpoint is

(-1-3)/2=-2

The y-coordinate of the vertex is -4 (because the range is [−4, ∞))

therefore

The vertex of f(x) is the point (-2,-4)

Is a vertical parabola open upward

The vertex is a minimum

The domain is all real numbers

step 2

we have

$g(x)=2x^2+8x+6$

Find the vertex

Factor the leading coefficient

$g(x)=2(x^2+4x)+6$

Complete the square

$g(x)=2(x^2+4x+4)+6-8$

$g(x)=2(x^2+4x+4)-2$

Rewrite as perfect squares

$g(x)=2(x+2)^2-2$

Is a vertical parabola open upward

The vertex is the point (-2,-2)

The domain is all real numbers

The range is the interval [−2, ∞)

Find the x-intercepts

For g(x)=0

$0=2(x+2)^2-2$

$2(x+2)^2=2$

$(x+2)^2=1\\x+2=\pm1\\x=-2\pm1$

The roots or x-intercepts are

x=-1 and x=-3

<u><em>Verify each statement</em></u>

A) Both functions have the same vertex

The statement is false

The vertex of f(x) is (-2,-4) and the vertex of g(x) is (-2,-2)

B) Both functions have the same domain.

The statement is true

The domain is all real numbers

C) Both functions have the same x-intercepts

The statement is true (see the explanation)

D) Both functions are decreasing on the interval (−∞, −2)

The statement is true

Because the x-coordinate of the vertex is the same in both functions, and both functions open upward

7 0

3 years ago

Read 2 more answers

If (P-3) is the largest of for odd number then the smallest is

zmey [24]

ensalada no son vulgares

6 0

2 years ago

What is the center of the circle described by the equation (x - 2)^2 +(y + 7)^2 =169

Marizza181 [45]

Answer:

The center is (2, -7)

Step-by-step explanation:

A circle is given by

(x-h)^2 + (y-k)^2 = r^2

where(h,k) is the center and r is the radius

(x - 2)^2 +(y + 7)^2 =169

(x - 2)^2 +(y - -7)^2 =13^2

The center is (2, -7) and the radius is 13

4 0

2 years ago

Which of the following statements are true about the graph of the function f(x)=(x+5)(x-3)

mina [271]

x * x = x^2

x * -3 = -3x

5 * x = 5x

5 * -3 = -15

x^2 -3x + 5x -15

x^2 + 2x -15

This function is a quadratic function, with an upwards parabola.

4 0

2 years ago