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vazorg [7]
2 years ago
7

Use the distributive property (expand) to write an equivalent expression for: 3(4x-2t). Type your answer ( I NEED THIS RIGHT AWA

Y PLS)
Mathematics
1 answer:
Eddi Din [679]2 years ago
5 0

Answer:

The distributive property here is to just multiply each value by tree number outside the parenthesis.

3 ( 4x - 2t)

12x - 6t

That would be your answer.

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How much interest will be paid in 1 year for a loan of $3600 at 8 1/2% simple interest?
solong [7]
The answer to your question is 260,100
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3 years ago
A food company sells peas in a can that has a radius of 3.4 cm and a height of 6 cm. Which is the volume of the can? (Use 3.14 f
Tanzania [10]

Answer:

217.9 cm

Step-by-step explanation:

V=πr2h=π·3.42·6≈217.90087

4 0
3 years ago
What is 9x+12y<br> ._. <br> so can you help me ?? ...umm :p
Viefleur [7K]

Simplifying

9x = 12y

Solving

9x = 12y

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Divide each side by '9'.

x = 1.333333333y

Simplifying

x = 1.333333333y

6 0
3 years ago
Will someone please help with this please ASAP
geniusboy [140]
A nice, interesting question. We have to be known to a equation called as the Circle equation. It is given by the formula of:

\boxed{\mathbf{(x - a)^2 + (y - b)^2 = r^2}}

That is the circle equation with a representation of the variable "a" and variable "b" as the points for the circle's center and the variable of "r" is representing the radius of the circle.

We are told to convert the given equation expression into a typical standard format of circle equation. This will mean we can easily deduce the values of the following variables and/or the points of the circle including the radius of the circle by our standard circle equation via conversion of this expression. So, let us start by interpreting this through equation editor for mathematical expression LaTeX, for a clearer view and better understanding.

\boxed{\mathbf{Given \: \: Equation: x^2 + y^2 - 4x + 6y + 9 = 0}}

Firstly, shifting the real numbered values or the loose number, in this case it is "9", to the right hand side, since we want an actual numerical value and the radius of circle without complicating and stressing much by using quadratic equations. So:

\mathbf{x^2 - 4x + 6y + y^2 = - 9}

Group up the variables of "x" and "y" for easier simplification.

\mathbf{\Big(x^2 + 4x \Big) + \Big(y^2 + 6y \Big) = - 9}

Here comes the catch of applying logical re-squaring of variables. We have to convert the variable of "x" into a "form of square". We can do this by adding up some value on the grouped variables as separately for "x" and "y" respectively. And add the value of "4" on the right hand side as per the square conversion. So:

\mathbf{\Big(x^2 - 4x + 4 \Big) + \Big(y^2 + 6y \Big) = - 9 + 4}

We can see that; our grouped variable of "x" is exhibiting the square of expression as "(x - 2)^2" which gives up the same expression when we square "(x - 2)^2". Put this square form back into our current Expressional Equation.

\mathbf{(x - 2)^2 + \Big(y^2 + 6y \Big) = - 9 + 4}

Similarly, convert the grouped expression for the variable "y" into a square form by adding the value "9" to grouped expression of variable "y" and adding the same value on the right hand side of the Current Equation, as per the square conversion.

\mathbf{(x - 2)^2 + \Big(y^2 + 6y + 9 \Big) = - 9 + 4 + 9}

Again; We can see that; our grouped variable of "y" is exhibiting the square of expression as "(y + 3)^2" which gives up the same expression when we square "(y + 3)^2". Put this square form back into our current Expressional Equation.

\mathbf{(x - 2)^2 + (y + 3)^2 = - 9 + 13}

\mathbf{(x - 2)^2 + (y + 3)^2 = 4}

Re-configure this current Expressional Equational Variable form into the current standard format of Circle Equation. Here, "(y - b)^2" is to be shown and our currently obtained Equation does not exhibit that. So, we do just one last thing. We distribute the parentheses and apply the basics of plus and minus rules. That is, "- (- 3)" is same as "+ (3)". And "4" as per our Circle Equation can be re-written as a exponential form of "2^2"

\mathbf{(x - 2)^2 + \big(y - (- 3) \big)^2 = 2^2}

Compare this to our original standard form of Circle Equation. Here, the center points "a" and "b" are "2" and "- 3". The radius is on the right hand side, that is, "2".

\boxed{\mathbf{\underline{\therefore \quad Center \: \: (a, \: b) = (2, \: - 3); \: Radius \: \: r = 2}}}

Hope it helps.
5 0
3 years ago
Witch point in a triangle is equidistant form the vertices of the triangle
Dafna1 [17]
It’s the circumcenter, or the center of the circle in which the triangle is inscribed.
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