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3 years ago

State the range of the numbers displayed in column x and column y. xy 15 2739

1 answer:

6 0

We can not really tell in this question as you dont know the equation that is being used for the domain and range relationship but overall one should know that:

The set of values of the independent variable(s) for which a function or relation is defined as the domain of a function. Typically, this is the set of x-values that give rise to real y-values.

The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain.

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Detmermine the best method to solve the following equation, then solve the equation. (3x-5)^2=-125

liq [111]

For the given equation;

$(3x-5)^2=-125$

We shall begin by expanding the parenthesis on the left side, after which we would combine all terms on and move all of them to the left side, which shall yield a quadratic equation. Then we shall solve.

Let us begin by expanding the parenthesis;

$\begin{gathered} (3x-5)^2\Rightarrow(3x-5)(3x-5) \\ (3x-5)(3x-5)=9x^2-15x-15x+25 \\ (3x-5)^2=9x^2-30x+25 \end{gathered}$

Now that we have expanded the left side of the equation, we would have;

$\begin{gathered} 9x^2-30x+25=-125 \\ \text{Add 125 to both sides and we'll have;} \\ 9x^2-30x+25+125=-125+125 \\ 9x^2-30x+150=0 \end{gathered}$

We shall now solve the resulting quadratic equation using the quadratic formula as follows;

$\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{Where;} \\ a=9,b=-30,c=150 \\ x=\frac{-(-30)\pm\sqrt[]{(-30)^2-4(9)(150)}}{2(9)} \\ x=\frac{30\pm\sqrt[]{900-5400}}{18} \\ x=\frac{30\pm\sqrt[]{-4500}}{18} \\ x=\frac{30\pm\sqrt[]{-900\times5}}{18} \\ x=\frac{30\pm\sqrt[]{-900}\times\sqrt[]{5}}{18} \\ x=\frac{30\pm30i\sqrt[]{5}}{18} \\ \text{Therefore;} \\ x=\frac{30+30i\sqrt[]{5}}{18},x=\frac{30-30i\sqrt[]{5}}{18} \\ \text{Divide all through by 6, and we'll have;} \\ x=\frac{5+5i\sqrt[]{5}}{3},x=\frac{5-5i\sqrt[]{5}}{3} \end{gathered}$

ANSWER:

$x=\frac{5+5i\sqrt[]{5}}{3},x=\frac{5-5i\sqrt[]{5}}{3}$

3 0

1 year ago

The length of each side of an equilateral

Svetlanka [38]

Answer:

6in

it would be 6inches 2sides or 12 an one be 6

6 0

3 years ago

2/5 of a foot pieces are there in 10 feet?

prohojiy [21]

Answer:

$\frac{a}{25}$ That is the answer if you're finding the exact form...

Step-by-step explanation:

Simplify the expression

5 0

3 years ago

I dont understand this at all. I really need help!

Westkost [7]

Hello!

$\large\boxed{x^{4}}$

Recall that:

$\sqrt[z]{x^{y} }$ is equal to $x^{\frac{y}{z} }$ . Therefore:

$\sqrt[3]{x^{2} } = x^{\frac{2}{3} }$

There is also an exponent of '6' outside. According to exponential properties, when an exponent is within an exponent, you multiply them together. Therefore:

$(x^{\frac{2}{3} })^{6} = x^{\frac{2}{3}* 6 } = x^{\frac{12}{3} } = x^{4}$

3 0

3 years ago

Can you please help me solve and if you show work I would really appreciate it

Anettt [7]

You got the equations correct, great job on that!

Let "s" be the variable that represents how many shirts were bought. Let "p" represent the total price/cost.

Equation for the store at Town Center mall:

p = 80 + 3.5s (80 is base cost, and cost increases 3.5 per shirt)

Equation for the store in Arlington:

p = 120 + 2.5s (120 is base cost, and cost increases 2.5 per shirt)

We want to find a point where the systems are equal; thus, we are solving for a system of linear equations, and we already have the equations we need.

p = 80 + 3.5s

p = 120 + 2.5s

We know that variable "p" is equal for both equations; thus, we can combine both equations into:

80 + 3.5s = 120 + 2.5s

Subtract both sides by 2.5s

80 + 3.5s - 2.5s = 120 + 2.5s - 2.5s

80 + s = 120

Subtract both sides by 80

s = 40

Thus, both equations are equal when 40 shirts are bought.

To find the cost, use any of the two equations (or both) to find the total cost, which should be equal.

p = 80 + 3.5(40) = 220

p = 120 + 2.5(40) = 220

Thus, the total price/cost at both stores is $220.

Let me know if you need any clarifications, thanks!

8 0

3 years ago