SAT Math - New SAT Math - No Calculator

Simplify the radical expression.

$\sqrt[3]{-64x^{6}y^{12}}$

$-4x^{2}y^{4}$

Explanation

$\sqrt[3]{-64x^{6}y^{12}}$

Look for perfect cubes within each term. This will allow us to factor out of the radical.

$\sqrt[3]{-64}*\sqrt[3]{x^6}*\sqrt[3]{y^{12}}$

$\sqrt[3]{-4*-4*-4}*\sqrt[3]{x^2*x^2*x^2}*\sqrt[3]{y^4*y^4*y^4}$

Simplify.

$-4x^{2}y^4$

Simplify the radical expression.

$\sqrt[3]{-64x^{6}y^{12}}$

$-4x^{2}y^{4}$

Explanation

$\sqrt[3]{-64x^{6}y^{12}}$

Look for perfect cubes within each term. This will allow us to factor out of the radical.

$\sqrt[3]{-64}*\sqrt[3]{x^6}*\sqrt[3]{y^{12}}$

$\sqrt[3]{-4*-4*-4}*\sqrt[3]{x^2*x^2*x^2}*\sqrt[3]{y^4*y^4*y^4}$

Simplify.

$-4x^{2}y^4$

Simplify the radical expression.

$\sqrt[3]{-64x^{6}y^{12}}$

$-4x^{2}y^{4}$

Explanation

$\sqrt[3]{-64x^{6}y^{12}}$

Look for perfect cubes within each term. This will allow us to factor out of the radical.

$\sqrt[3]{-64}*\sqrt[3]{x^6}*\sqrt[3]{y^{12}}$

$\sqrt[3]{-4*-4*-4}*\sqrt[3]{x^2*x^2*x^2}*\sqrt[3]{y^4*y^4*y^4}$

Simplify.

$-4x^{2}y^4$

Simplify the radical expression.

$\sqrt[3]{-64x^{6}y^{12}}$

$-4x^{2}y^{4}$

Explanation

$\sqrt[3]{-64x^{6}y^{12}}$

Look for perfect cubes within each term. This will allow us to factor out of the radical.

$\sqrt[3]{-64}*\sqrt[3]{x^6}*\sqrt[3]{y^{12}}$

$\sqrt[3]{-4*-4*-4}*\sqrt[3]{x^2*x^2*x^2}*\sqrt[3]{y^4*y^4*y^4}$

Simplify.

$-4x^{2}y^4$

Simplify the radical expression.

$\sqrt[3]{-64x^{6}y^{12}}$

$-4x^{2}y^{4}$

Explanation

$\sqrt[3]{-64x^{6}y^{12}}$

Look for perfect cubes within each term. This will allow us to factor out of the radical.

$\sqrt[3]{-64}*\sqrt[3]{x^6}*\sqrt[3]{y^{12}}$

$\sqrt[3]{-4*-4*-4}*\sqrt[3]{x^2*x^2*x^2}*\sqrt[3]{y^4*y^4*y^4}$

Simplify.

$-4x^{2}y^4$

What is the sum of all the values of that satisfy:

$\frac{7}{3}$

$\frac{2}{3}$

$\frac{14}{15}$

Explanation

With quadratic equations, always begin by getting it into standard form:

Therefore, take our equation:

And rewrite it as:

You could use the quadratic formula to solve this problem. However, it is possible to factor this if you are careful. Factored, the equation can be rewritten as:

Now, either one of the groups on the left could be and the whole equation would be . Therefore, you set up each as a separate equation and solve for :

$x=\frac{-2}{3}$

The sum of these values is:

$3+(-\frac{2}{3})=3-\frac{2}{3}=\frac{9}{3}-\frac{2}{3}=\frac{7}{3}$

What is the sum of all the values of that satisfy:

$\frac{7}{3}$

$\frac{2}{3}$

$\frac{14}{15}$

Explanation

With quadratic equations, always begin by getting it into standard form:

Therefore, take our equation:

And rewrite it as:

You could use the quadratic formula to solve this problem. However, it is possible to factor this if you are careful. Factored, the equation can be rewritten as:

Now, either one of the groups on the left could be and the whole equation would be . Therefore, you set up each as a separate equation and solve for :

$x=\frac{-2}{3}$

The sum of these values is:

$3+(-\frac{2}{3})=3-\frac{2}{3}=\frac{9}{3}-\frac{2}{3}=\frac{7}{3}$

What is the sum of all the values of that satisfy:

$\frac{7}{3}$

$\frac{2}{3}$

$\frac{14}{15}$

Explanation

With quadratic equations, always begin by getting it into standard form:

Therefore, take our equation:

And rewrite it as:

You could use the quadratic formula to solve this problem. However, it is possible to factor this if you are careful. Factored, the equation can be rewritten as:

Now, either one of the groups on the left could be and the whole equation would be . Therefore, you set up each as a separate equation and solve for :

$x=\frac{-2}{3}$

The sum of these values is:

$3+(-\frac{2}{3})=3-\frac{2}{3}=\frac{9}{3}-\frac{2}{3}=\frac{7}{3}$

What is the sum of all the values of that satisfy:

$\frac{7}{3}$

$\frac{2}{3}$

$\frac{14}{15}$

Explanation

With quadratic equations, always begin by getting it into standard form:

Therefore, take our equation:

And rewrite it as:

You could use the quadratic formula to solve this problem. However, it is possible to factor this if you are careful. Factored, the equation can be rewritten as:

Now, either one of the groups on the left could be and the whole equation would be . Therefore, you set up each as a separate equation and solve for :

$x=\frac{-2}{3}$

The sum of these values is:

$3+(-\frac{2}{3})=3-\frac{2}{3}=\frac{9}{3}-\frac{2}{3}=\frac{7}{3}$

What is the sum of all the values of that satisfy:

$\frac{7}{3}$

$\frac{2}{3}$

$\frac{14}{15}$

Explanation

With quadratic equations, always begin by getting it into standard form:

Therefore, take our equation:

And rewrite it as:

You could use the quadratic formula to solve this problem. However, it is possible to factor this if you are careful. Factored, the equation can be rewritten as:

Now, either one of the groups on the left could be and the whole equation would be . Therefore, you set up each as a separate equation and solve for :

$x=\frac{-2}{3}$

The sum of these values is:

$3+(-\frac{2}{3})=3-\frac{2}{3}=\frac{9}{3}-\frac{2}{3}=\frac{7}{3}$

New SAT Math - No Calculator

Help Questions

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation