Answer:
![\boxed {\boxed {\sf \sqrt {145} \ or \ 12.04}}](https://img.qammunity.org/2022/formulas/mathematics/college/o53fu5874g1t7b5cmaqjlh4wfugfrlt5qj.png)
Explanation:
The distance between 2 points is calculated using the following formula.
![d= \sqrt {(x_2-x_1)^2+(y_2-y_1)^2)](https://img.qammunity.org/2022/formulas/mathematics/college/dahi7v2qxui93v31sypj42jqpghcysxk4j.png)
In this formula, (x₁, y₁) and (x₂, y₂) are the coordinates of the two points.
We know the two points are (4, -9) and (5,3). If we match the values of the points and the coordinating variable, we see that:
- x₁ = 4
- y₁= -9
- x₂ = 5
- y₂ = 3
Substitute the values into the formula.
![d= \sqrt { ( 5 -4)^2 + ( 3 --9)^2](https://img.qammunity.org/2022/formulas/mathematics/college/kislnajr7cbbn7yj8hnx27sb9i41ccq191.png)
Solve inside the parentheses.
- (5-4)= 1
- (3 --9) = (3+9) = 12
![d= \sqrt {(1)^2 + (12)^2}](https://img.qammunity.org/2022/formulas/mathematics/college/njv2kl7cejgkmfholyjj2231qjjozmnd0p.png)
Solve the exponents.
- (1)² = 1 *1 = 1
- (12)² = 12 * 12 = 144
![d= √( 1+144)](https://img.qammunity.org/2022/formulas/mathematics/college/f4hv8u9mvgcq1wn8g8j7qosa0wq83b4svr.png)
Add.
![d= \sqrt{145](https://img.qammunity.org/2022/formulas/mathematics/college/ephc9wltulccijzq1vg7lko55ovxkbne7n.png)
Take the square root.
![d=12.04159458](https://img.qammunity.org/2022/formulas/mathematics/college/ra94pb474jvm8vvn551w9zfxgp4lzjbtiw.png)
Let's round to the nearest hundredth. The 1 in the thousandth place tells us to leave the 4 in the hundredth place.
![d \approx 12.04](https://img.qammunity.org/2022/formulas/mathematics/college/j1k79m2f0u2ham4zc1qu6vckxlwv5tmeyv.png)
The distance between the 2 points is √145 or approximately 12.04.