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jeka57 [31]
3 years ago
9

Im on a time limit please help!!!!

Mathematics
2 answers:
vaieri [72.5K]3 years ago
7 0

Answer:

D

Step-by-step explanation:

denpristay [2]3 years ago
5 0

Answer:

Option D

Step-by-step explanation:

Denote 2 tangent points are A and B

=> angle ACB = arc AB = 164 degree

We have:

ACB + x = 180

=> x = 180 - ACB = 180 - 164 = 16 degree

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Find the x-intercept for y = 3x^2 + 6x − 10
azamat

Answer:

x-intercepts: (-3.08, 0) and (1.08, 0)

Step-by-step explanation:

Given:

The function is given as:

y=3x^2+6x-10

In order to find the x-intercept, we need to equate the given function to 0 as x-intercept is the point where the 'y' value is 0. So,

y=0\\3x^2+6x-10=0

Now, this is a quadratic equation of the form ax^2+bx+c=0

We find the solution using the quadratic formula,

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

Here, a=3,b=6,c=-10

Now, the solutions are:

x=\frac{-6\pm \sqrt{6^2-4(3)(-10)}}{2(3)}\\\\x=\frac{-6\pm \sqrt{36+120}}{6}\\\\x=\frac{-6\pm \sqrt{156}}{6}\\\\x=\frac{-6}{6}-\frac{2\sqrt{39}}{6}\ or\ x=\frac{-6}{6}+\frac{2\sqrt{39}}{6}\\\\x=-1-2.08\ or\ x=-1+2.08\\\\x=-3.08\ or\ x=1.08

Therefore, the x-intercepts are (-3.08, 0) and (1.08, 0)

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