According to the secant-tangent theorem, we have the following expression:
Now, we solve for <em>x</em>.
Then, we use the quadratic formula:
![x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x_%7B1%2C2%7D%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D)
Where a = 1, b = 6, and c = -315.
![\begin{gathered} x_{1,2}=\frac{-6\pm\sqrt[]{6^2-4\cdot1\cdot(-315)}}{2\cdot1} \\ x_{1,2}=\frac{-6\pm\sqrt[]{36+1260}}{2}=\frac{-6\pm\sqrt[]{1296}}{2} \\ x_{1,2}=\frac{-6\pm36}{2} \\ x_1=\frac{-6+36}{2}=\frac{30}{2}=15 \\ x_2=\frac{-6-36}{2}=\frac{-42}{2}=-21 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x_%7B1%2C2%7D%3D%5Cfrac%7B-6%5Cpm%5Csqrt%5B%5D%7B6%5E2-4%5Ccdot1%5Ccdot%28-315%29%7D%7D%7B2%5Ccdot1%7D%20%5C%5C%20x_%7B1%2C2%7D%3D%5Cfrac%7B-6%5Cpm%5Csqrt%5B%5D%7B36%2B1260%7D%7D%7B2%7D%3D%5Cfrac%7B-6%5Cpm%5Csqrt%5B%5D%7B1296%7D%7D%7B2%7D%20%5C%5C%20x_%7B1%2C2%7D%3D%5Cfrac%7B-6%5Cpm36%7D%7B2%7D%20%5C%5C%20x_1%3D%5Cfrac%7B-6%2B36%7D%7B2%7D%3D%5Cfrac%7B30%7D%7B2%7D%3D15%20%5C%5C%20x_2%3D%5Cfrac%7B-6-36%7D%7B2%7D%3D%5Cfrac%7B-42%7D%7B2%7D%3D-21%20%5Cend%7Bgathered%7D)
<h2>Hence, the answer is 15 because lengths can't be negative.</h2>
Answer:
The other number is 2 if the given number is 15.
Step-by-step explanation:
One of the numbers is 15, correct?
Let the other number be x.
15 × x = 30
x = 30/15
x = 2.
Let's check:
15 × 2 = 30.
30 = 30.
Correct!
In function notation, where f is the function, f(x)=3x-10. Using f(x) means you can write things like f(0) which means: what is the value of the function when x=0? Think of f for flexibility, because functions have greater flexibility than variables like y.
Answer:
89
Step-by-step explanation:
88+190/2=89 I hope this is enough
