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2 years ago

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The red figure is the transformed image of the blue figure. Gloria correctly identifies a single transformation that can move th

e blue figure onto the red figure. Layla correctly identifies a sequence of two transformations that can be used. Use the drop-down menus below to complete the statements describing the transformations they used.

1 answer:

julia-pushkina [17]2 years ago

5 0

Answer:

The answers are;

First the blue image is translated -2x to the second quadrant quadrant and image (-x, y)

Then it is rotated 180° to get the the red figure with coordinates (x, -y)

Step-by-step explanation:

Two transformation that is equivalent to a reflection

We note that the x-coordinate values remain the same while the y-coordinate values change from +y to -y

Therefore, the two transformations are;

First the blue image with coordinates (x, y) is translated by -2x using the transformation T(-2x, 0) to get (-x, y)

Then it is rotated 180° to get the point (x, -y) which is the coordinate of the pink image.

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When is it easiest to find the percent of a number using a decimal??

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I believe when it's a fraction.

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3 years ago

Which equation can be used to find A, the area of the parallelogram shown?

Korvikt [17]

Answer:

A = 6 x 7

Step-by-step explanation:

To find the area of a parallelogram we use the equation of Area = base x height. In this case since the base is 6 and the height is 7, we plus in these numbers for the variables (b x h) to get the area.

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3 years ago

A 15 kilogram object is suspended from the end of a vertically hanging spring stretches the spring 1/3 meters. At time t = 0, th

Yuri [45]

Answer:

$15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) =$ ± $170 cos(5t)$

$y(0)=0$ , $y'(0)=0$

Step-by-step explanation:

See the attached image

This problem involves Newton's 2nd Law which is: ∑F = ma, we have that the acting forces on the mass-spring system are: $F_{r} (t)$ that correspond to the force of resistance on the mass by the action of the spring and $F(t)$ that is an external force with unknown direction (that does not specify in the enounce).

For determinate $F_{r} (t)$ we can use Hooke's Law given by the formula $F_{r} (t) = k y(t)$ where $k$ correspond to the elastic constant of the spring and $y(t)$ correspond to the relative displacement of the mass-spring system with respect of his rest state.

We know from the problem that an 15 Kg mass stretches the spring 1/3 m so we apply Hooke's law and obtain that...

$k = \frac{F_{r}}{y} = \frac{mg}{y} = \frac{15 Kg (9.81 \frac{m}{s^{2} } )}{\frac{1}{3} m} = 441.45 \frac{N}{m}$

Now we apply Newton's 2nd Law and obtaint that...

$F_{r} (t)$ ± $F(t)$ = $ma(t)$

$F_{r} (t) = ky(t) = 441.45y(t)$

$F(t) = 170 cos(5t)$

$m = 15 kg$

$a(t) = \frac{d^{2}y(t)}{dt^{2} }$

Finally... $15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) =$ ± $170 cos(5t)$

We know from the problem that there's not initial displacement and initial velocity, so... $y(0)=0$ and $y'(0)=0$

Finally the Initial Value Problem that models the situation describe by the problem is

$\left \{ 15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) = \frac{+}{} 170 cos(5t) \atop {y(0)=0, y'(0)=0\right.$

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2 years ago

28 more than a number ?

siniylev [52]

Answer:

28 + x

Step-by-step explanation:

To write this as an expression, you need to find out what sign you need to use ( multiplication, division,etc.)

More than means to add so you are using an addition sign. Now you need a variable to represent the number.

You can pick any letter but for this I am using x. When you put that all together you get the expression, 28 + x.

4 0

3 years ago

The area of a triangle is 15x^4+3x^3+4x^2-x-3 square meters. The length of the base of the triangle is 6x^2-2 meters. What is th

MArishka [77]

Answer:

The height of the triangle is: $H = (5x^2 + x +3)$

Step-by-step explanation:

Here, the area the triangle is given as: $15x^4 + 3x^3 + 4x^2 - x - 3$

Also, base of the triangle = $(6x^2 -2)$

Let us assume the height of the triangle = H units

Now, AREA of TRIANGLE = $\frac{1}{2} \times B \times H$

$\implies (15x^4+3x^3+4x^2-x-3 ) = \frac{1}{2} \times ( 6x^2-2) \times H\\\implies H =\frac{ (15x^4+3x^3+4x^2-x-3 )}{(3x^2-1)}$

Solving by LONG DIVISION, we get:

$H = (5x^2 + x +3)$

Hence, the height of the triangle is: $H = (5x^2 + x +3)$

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3 years ago