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3 years ago

PLEASE HELP!!

2 answers:

5 0

Standard form is just putting (in this case) variables in alphabetical order. First we simplify- 2x+4=4y would become 1/2x+1=y. This is already in standard form, as numbers w/o variables come at the end. simplifying the next one is more tricky. first you get a variable/number alone-
9-2x-2y=4x+3
9-2y=6x+3
6-2y=6x
1-1/3y=x
1=x+1/3y
3=x+y
x+y=3
sorry if I was wrong and not of any help, But I do believe this is correct.

lbvjy [14]3 years ago

3 0

What standard for is is when all the variables are on one side in order, without being expressed as a decimal or fraction.

To put 2x+4=4y in standard form, first subtract 4 from both sides.

2x=4y-4
Now, subtract 4y from both sides.

2x-4y=-4

Now the equation is in standard form.

To put the next equation in order, first distribute -2 into (x-y) to get:

9-2x+2y=4x+3

Because one variable cannot be on both sides, subtract 4x from both sides.

9-6x+2y=3

Now subtract 9 to both sides:

-6x+2y=-6

My teacher last year did not like when the first variable in standard form was negative. If that is the case with your teacher, distribute -1 into everything on both sides of the equation. That would look like:

6x-2y=6

If your teacher is impartial, then you can leave there equation how it was.

So the answers are:

2x-4y=-4
-6x+2y=-6 (or 6x-2y=6)

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3 years ago

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Help me please ;-;!!!!! Use a coordinate grid to create a map of a town with at least five different locations, such as a house,

Sholpan [36]

A. Our diagram have six location: my house (2,1), a movie theater (5,3), the school (6,6), the police station (3,-2), the airport (-3,3), and the mall (-2,-3)

B. Distance between my house and the movie theater:
To calculate the distance we are going to use the points (2,1) and (5,3) to create a right triangle with tow legs of measures 3 and 2; the hypotenuse of the right triangle will be the distance between my house and the movie theater.
Using the Pythagorean theorem:
$d^2=3^2+2^2$
$d^2=9+4$
$d^2=13$
$d= \sqrt{13}$
We can conclude that the distance between my house and the movie theater is $\sqrt{13}$

Our second distance is the distance between the police station and the airport:
This time we are using the points (3,-2) and (-3,3) to create a right triangle with legs of measures 6 and 5; the hypotenuse of our triangle will be the the distance between our points:
$d^2=6^2+5^2$
$d^2=36+25$
$d^2=61$
$d= \sqrt{61}$
We can conclude that the distance between the police station and the airport is $\sqrt{61}$

C. Distance form the park to the Fire Station:
$d^2=3^2+1^2$
$d^2=9+1$
$d^2=10$
$d= \sqrt{10}$
We can conclude that the distance from the park to the Fire Station is $\sqrt{10}$

Distance from the stadium to the mall:
$d^2=2^2+1^2$
$d^2=4+1$
$d^2=5$
$d= \sqrt{5}$
We can conclude that the distance from the stadium to the mall is $\sqrt{5}$

D. What is the distance between your house and the mall?
Answer:
$d^2=4^2+4^2$
$d^2=16+16$
$d^2=32$
$d= \sqrt{32}$
$d=4 \sqrt{2}$
We can conclude that the distance between my house and the mall is $4 \sqrt{2}$

What is the distance between the movie theater and the school?
answer:
$d^2=3^2+1^2$
$d^2=9+1$
$d^2=10$
$d= \sqrt{10}$
We can conclude that the distance between the movie theater and the school is $\sqrt{10}$

7 0

3 years ago

the line y=2x-4 is dilated bt a scale factor of 3/2 and centered at the origin. Write an equation that represent that image of t

Veseljchak [2.6K]

Answer: $y = 2x -6$

Step-by-step explanation:

Since, If (x,y) represents the coordinates of a line, then after dilation about the origin by the scale factor k, the coordinates of the resultant line will get by the rule,

$(x,y)\rightarrow (kx,ky)$

Here, The equation of the given line is,

y = 2x - 4

x-intercept and y-intercept of the line are (2,0) and (0,-4) respectively,

Hence, the line y = 2x - 4 will be passes through the points (2,0) and (0,-4),

By the above property,

The dilated line about the origin will be passes through the points (3,0) and (0,-6), ( Because (3/2× 2, 3/2×0) = (3,0) and (3/2×0, 3/2×-4) = (0,-6) )

Hence, the equation of transformed line after dilation,

$y-0 = \frac{-6-0}{0-3}(x-3)$