How many milliliters are in 3 quarts

This is the question can someone help

agasfer [191]

Answer:

The coordinates of the vertices of triangle Q'R'S' will be:

Q'(-6, 9)

R'(-9, 3)

S'(-2, 2)

Step-by-step explanation:

Given the coordinates of the vertices QRS

The coordinates of the vertex Q is (-2, 3)

The coordinates of the vertex R is (-5, -3)

The coordinates of the vertex Q is (2, -4)

As we know the rules of translation that:

If you move horizontally 'c' units to the left, then 'c' units are subtracted to the x-coordinate of each of the vertices.

if you move 'c' units up, then 'c' units are added to the y-coordinate of each of the vertices.

Translation 6 units up and 4 units to the left.

(x, y) → (x - 4, y + 6)

As the triangle Q'R'S' is the image of the triangle QRS translated 6 units up and 4 units to the left.

So,

the coordinates of Q' will be:

(x, y) → (x - 4, y + 6) = (x, y) → (-2 - 4, 3 + 6) = (-6, 9)

the coordinates of R' will be:

(x, y) → (x - 4, y + 6) = (x, y) → (-5 - 4, -3 + 6) = (-9, 3)

the coordinates of S' will be:

(x, y) → (x - 4, y + 6) = (x, y) → (2 - 4, -4 + 6) = (-2, 2)

Therefore, the coordinates of the vertices of triangle Q'R'S' will be:

Q'(-6, 9)

R'(-9, 3)

S'(-2, 2)

6 0

3 years ago

Identify the error in this column:<br> (x2 + 3x + 18) = (x + 3)(x + 6)

inessss [21]

Answer:

I think the error is that the variable x came before the whole number 2.

3 0

3 years ago

Read 2 more answers

Why are triangles a special polygon?

Len [333]

<span> A triangle is the simplest polygon, having three sides and three angles. The sum of the three angles of a triangle is equal to 180 degrees. </span>

6 0

3 years ago

What will happen to the graph of the line y = -54x + 3 if you change -54 to 54?

viva [34]

Answer:

Graph the line using the slope and y-intercept, or two points.

Slope:

1

y-intercept: ( 0 , 0 ) x y 0 0 1 1

6 0

4 years ago

A rock thrown vertically upward from the surface of the moon at a velocity of 36m/sec reaches a height of s = 36t - 0.8 t^2 met

Verdich [7]

Answer:

a. The rock's velocity is $v(t)=36-1.6t \:{(m/s)}$ and the acceleration is $a(t)=-1.6 \:{(m/s^2)}$

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

Velocity is defined as the rate of change of position or the rate of displacement. $v(t)=\frac{ds}{dt}$
Acceleration is defined as the rate of change of velocity. $a(t)=\frac{dv}{dt}$

a.

The rock's velocity is the derivative of the height function $s(t) = 36t - 0.8 t^2$

$v(t)=\frac{d}{dt}(36t - 0.8 t^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(36t\right)-\frac{d}{dt}\left(0.8t^2\right)\\\\v(t)=36-1.6t$

The rock's acceleration is the derivative of the velocity function $v(t)=36-1.6t$

$a(t)=\frac{d}{dt}(36-1.6t)\\\\a(t)=-1.6$

b. The rock will reach its highest point when the velocity becomes zero.

$v(t)=36-1.6t=0\\36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\\360-16t=0\\360-16t-360=0-360\\-16t=-360\\t=\frac{45}{2}=22.5$

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

$s(22.5) = 36(22.5) - 0.8 (22.5)^2\\s(22.5) =405$

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height = $\frac{405}{2} =202.5 \:m$

To find the time it reach half its maximum height, we need to solve

$36t - 0.8 t^2=202.5\\36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\\360t-8t^2=2025\\360t-8t^2-2025=2025-2025\\-8t^2+360t-2025=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\\\\t=\frac{-360+\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2-\sqrt{2}\right)}{4}\approx 6.59\\\\t=\frac{-360-\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2+\sqrt{2}\right)}{4}\approx 38.41$

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

$36t - 0.8 t^2=0\\\\\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\t=0,\:t=45$

The rock is aloft for 45 seconds.

5 0

4 years ago