We know that
If a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment. (Intersecting Secant-Tangent Theorem)
so
ST²=RT*QT
RT=7 in
QT=23+7-----> 30 in
ST²=7*30-----> 210
ST=√210-----> 14.49 in
the answer is
RT=14.49 in
Answer:
At approximately x = 0.08 and x = 3.92.
Step-by-step explanation:
The height of the ball is modeled by the function:
Where f(x) is the height after x seconds.
We want to determine the time(s) when the ball is 10 feet in the air.
Therefore, we will set the function equal to 10 and solve for x:
Subtracting 10 from both sides:
For simplicity, divide both sides by -1:
We will use the quadratic formula. In this case a = 16, b = -64, and c = 5. Therefore:
Substitute:
Evaluate:
Simplify the square root:
Therefore:
Simplify:
Approximate:
Therefore, the ball will reach a height of 10 feet at approximately x = 0.08 and x = 3.92.
Answer:
z
=
Step-by-step explanation:
Answer:
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The answer is
the first one is x= 18.98
a^2+b^2=c^2
8 cm^2+17 cm^2= x^2
64+289=x^2
√x^2= √353
x=18.89
<span><span>the second one is x= 18.7
a^2+b^2=c^2
7 m^2+ x^2= 20 m^2
49 m+x^2=400 m
-49 -49
</span></span>√x^2=√352
<span><span>x= 18.7
Hope this helps!</span></span>
