Simplify 6 cm : 1.5 m

A rhombus has coordinates A(-6, 3), B(-4, 4), C(-2, 3), and D(-4, 2). What are the coordinates of rhombus A′B′C′D′ after a 90° c

Liula [17]

We are given that the coordinates of the vertices of the rhombus are:
A(-6, 3)
B(-4, 4)
C(-2, 3)
D(-4, 2)

To solve this problem, we must plot this on a graphing paper or graphing calculator to clearly see the movement of the graph. If we transform this by doing a counterclockwise rotation, then the result would be:
A(-6, -3)
B(-4, -4)
C(-2, -3)
D(-4, -2)

And the final transformation is translation by 3 units left and 2 units down. This can still be clearly solved by actually graphing the plot. The result of this transformation would be:

A′(6, -8)
B′(7, -6)
C′(6, -4)
D′(5, -6)

7 0

3 years ago

Read 2 more answers

X - 5 > -10. give the solution

expeople1 [14]

X > -10 + 5
bring the five to the other side of the equation and change the sign in front
x > -5

5 0

2 years ago

Read 2 more answers

Does the point (0, 0) satisfy the equation y = 9x?

miskamm [114]

Answer:

yes it does

Step-by-step explanation:

because the equation y=9x does not have a y-intercept (all slopes come in the form y=mx+b -- it can be written differently though) and since there is no 'b' that means the y-intercept is 0. So whenever there is no y-intercept, the slope starts at 0.

6 0

2 years ago

The mass of the particles that a river can transport is proportional to the sixth power of the speed of the river. A certain riv

kari74 [83]

The river flows at 2 miles per hour. This is the speed of the current, or the speed at which the water itself is moving.

The mass of the particles moving in river is proportional to the sixth power of 2, or $2^6 = 64$ miles per hour. So, if x is the mass of the particle, then the speed is modeled by an inverse relationship $x * 2^6 = k$ , for some constant k.

For particles that are 15 times the usual mass, the speed of the river itself must increase for the particles to travel at the same speed as before. The mass of the particles is still proportional to the sixth power of the speed of the river, but the speed of the river can no longer be 2 miles per hour. It is some speed y, which we are solving for.

So, the key here is proportional. We can set up two proportions and cross multiply.

$\dfrac{x}{2^6} = \dfrac{15x}{y^6} \\ (x)(y^6)=(2^6)(15x)$

Cancelling our x's (because we don't need to solve for them), we have

$y^6 = 15(2^6) = 15(64)=960$

To solve for y, we must take the sixth root of both sides, or raise both sides to the one-sixth power.

$\sqrt[6]{y^6} = \sqrt[6]{960}$

Using a calculator, we see that $\sqrt[6]{960} = 3.14 \ miles \ per \ hour$ .

6 0

3 years ago

(6-d)(d^2-5+3D) please help me answer this hahaha

levacccp [35]

If you're simplifying the equation your answer is going to become -d^3 +3d^2 +23d -30

3 0

3 years ago