What is the solution 6x-4 =2(3x-2)

What is 4x-2 equal to??

dsp73

If you mean 4 times -2 thats -8 if you mean 4x-2 then you cant do that

7 0

3 years ago

How to do this? i feel dizzy + crying when do this

Talja [164]

#1:
a) Q(1,2); R(1,6); S(7,6)
b) The point of intersection of PR and QS is (4,4)

#2:
a) S(3,2)
b) T(5,3)
c) Area = 8 units^2

6 0

3 years ago

ray is buying some ginger roots to brew some fresh ginger ale. The price of the ginger roots is G G, and Ray has a coupon for 1

zalisa [80]

Answer: 0.9 is the decreasing factor.

Step-by-step explanation:

Here, The price of the ginger roots is G .

And, It there occurs the 10% off on this amount then,

The new price of the ginger is 90% of G.

That is, Final price of G = $\frac{90\times G}{100}$

= $0.9\times G$

= 0.9G

Since, 0.9 decreases the price of the ginger.

Therefore, It is called decreasing factor for the function which represents the price of the ginger.

3 0

3 years ago

Read 2 more answers

Prove that if the set of vectors {u1, u2, u3} is linearly dependent and u4 is any vector, then the set {u1, u2, u3, u4} is linea

Lesechka [4]

Answer with Step-by-step explanation:

We are given that the set of vectors ${u_1,u_2,u_3}$ is lineraly dependent set .

We have to prove that the set ${u_1,u_2,u_3,u_4}$ is linearly dependent .

Linearly dependent vectors : If the vectors $u_1,u_2.u_3,u_4$

are linearly dependent therefore the linear combination

$a_1u_1+a_2u_2+a_3u_3+a_4u_4=0$

Then ,there exit a scalar which is not equal to zero .

Let $a_1\neq0$ then the vector $u_1$ will be zero and remaining other vectors are not zero.

Proof:

When $u_1,u_2,u_3$ are linearly dependent vectors therefore, linear combination of vectors of given set

$a_1u_1+a_2u_2+a_3u_3=0$

By definition of linearly dependent vector

There exist a scalar which is not equal to zero.

Suppose $a_1\neq 0$ then $u_1=0$

The linear combination of the set ${u_1,u_2,u_3,u_4}$

$a_1u_1+a_2u_2+a_3u_3+a_4u_4=0$

When $a_1\neq0\; and\; u_1=0$

Therefore,the set ${u_1,u_2,u_3,u_4}$ is linearly dependent because it contain a vector which is zero.

Hence, proved .

5 0

3 years ago

Y = x² – 4 y = 2r + 4<br><br>Solve algebraicallu for the solutions of equations below.

oksano4ka [1.4K]

Step-by-step explanation:

Okay, the first step is to rewrite this equation in "vertex form." You can search that up but it's basically just (h/k).

y = ( x − 1) 2 + 3

Now, we are going to use the vertex form, "y = a (x - h)2 + k, to get the values of a, h, and k.

By using the form we get 1 for the value a, 1 for the value h, and 3 for the value k.

1 = a

1 = h

3 = k

Becuse the value of a is positive, the parabola opens up! (A parabola is the U shaped line in a graph, so that opens up.)

Now, we find the vertex (h,k)

which is (1,3)

Now we find the p from the vertex to the focus. (vertext the top, focus one of the points.)

Follow this formula to find the distance from the vertex to a focus by using this formula

1/4a.

Now we just gonna substitue 1 for a, since we know that 1 equals a above ^

1/4 * 1

Solving that, we got a nice little 1/4.

Next we find the focus

"Find the focus.

The focus of a parabola can be found by adding p to the y-coordinate k if the parabola opens up or down.

(h,k+p)

Substitute the known values of h, p, and k into the formula and simplify.

(1,134)

Find the axis of symmetry by finding the line that passes through the vertex and the focus.

x=1

Find the directrix.

y=114

Use the properties of the parabola to analyze and graph the parabola.

Direction: Opens Up

Vertex: (1,3)

Focus: (1,134)

Axis of Symmetry: x=1

Directrix: y=114

Select a few x values, and plug them into the equation to find the corresponding y values. The x values should be selected around the vertex.

xy−1704132437

Graph the parabola using its properties and the selected points.

Direction: Opens Up

Vertex: (1,3)

Focus: (1,134)

Axis of Symmetry: x=1

Directrix: y=114

xy−17041324"

(Sorry if this is long! But I hope you understand it better now! Thanks for the points!)

3 0

3 years ago