Answer:
![\boxed {\boxed {\sf x \approx 30.9}}](https://img.qammunity.org/2021/formulas/mathematics/middle-school/jwncozvlb8e6luw1s7zes55279rb6a2ykh.png)
Explanation:
This is a right triangle and we are looking for a missing side. We can use the Pythagorean Theorem to find x.
![a^2+b^2=c^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/96dopf217hvzc3zhswffnjr8l5f26vmjhb.png)
where a and b are the legs and c is the hypotenuse.
In this triangle, the legs are 28 and 13 because they make up the right right angle. x is the hypotenuse because it is opposite the right angle.
![a= 28 \\b= 13 \\c= x](https://img.qammunity.org/2021/formulas/mathematics/middle-school/qguweom8p8dk1ld0cibnboi3gujtpggpfq.png)
Substitute the values in.
![(28)^2+(13)^2=x^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/3pdlsflirwqorpjwudnxngm9vb3tfq9qd5.png)
Solve the exponents.
- (28)²=28*28=784
- (13)²=13*13=169
![784+169=x^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/c79qndrr2j32dtdfbxi7zvaluqjjpzjbg8.png)
Add the two numbers.
![{953} =x^2\\](https://img.qammunity.org/2021/formulas/mathematics/middle-school/9kfg6q2tyb0fe50jx0sux6srmfrx9937ox.png)
We want to solve for x, which means we must isolate it.
The x is being squared. The inverse of a square is the square root, so take the square root of both sides of the equation.
![√(953)=√(x^2) \\](https://img.qammunity.org/2021/formulas/mathematics/middle-school/yrm060qoi92h49ijhpfm7zkxrfthzx9sv2.png)
![√(953) = x](https://img.qammunity.org/2021/formulas/mathematics/middle-school/efuv8cxdc4vfs8w37ez1bhamd9nm4z22ek.png)
![30.87069808=x](https://img.qammunity.org/2021/formulas/mathematics/middle-school/jv6crzrfqq4janw27xh4mxowjl3lwgz45y.png)
Let's round to the nearest hundredth. The 7 in the thousandth place tells us to round the 8 to a 9.
![30.9 \approx x](https://img.qammunity.org/2021/formulas/mathematics/middle-school/z65ztfe4n1uejyole30bl04n6qq60x1pmh.png)
x is about 30.9