The answer will be D. 7/10 because you have 10 numbers and you want to have a number greater than 3 so it would be 10-3=7 and 7 would go over 10 because there are 7 numbers greater than 3 but less than 10.
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>
3.2 less days than 16 because each 25 is 5 than 16 divide by 5 is 3.2
Answer:
Look below
Step-by-step explanation:
lets have -29 be y2 and -5 be x2
21 - (-)29
5 - (-)5
Subtracting a negative is just adding it so
21 + 29 50
5 + 5 10
50/10
this simplified, is 5/1 or 5
y=5x+b
now we need to find b, so we substitute y and y
x = 5
y = 21
21=5*5=b
5*5 is 25
now we need to get to 21
25 -4 is 21
b = -4
y=5x-4
