4(x + 3) = 6 - x
First, expand to remove parentheses.
Second, subtract '6' from both sides.
Third, subtract '12 - 6' to get 6.
Fourth, subtract '4x' from both sides.
Fifth, since 'x' can be referred to as '1', add it to '4x' to get '-5x'.
Sixth, divide both sides by '-5'.
Seventh, change the whole fraction into a negative.
Eighth, switch your sides.
![x = -\frac{6}{5}](https://tex.z-dn.net/?f=x%20%3D%20%20-%5Cfrac%7B6%7D%7B5%7D%20)
Answer as fraction:
![-\frac{6}{5}](https://tex.z-dn.net/?f=%20-%5Cfrac%7B6%7D%7B5%7D%20)
Answer as decimal: -1.2
Answer:
Part 1) m∠1 =(1/2)[arc SP+arc QR]
Part 2) ![PR^{2} =PS*PT](https://tex.z-dn.net/?f=PR%5E%7B2%7D%20%3DPS%2APT)
Part 3) PQ=PR
Part 4) m∠QPT=(1/2)[arc QT-arc QS]
Step-by-step explanation:
Part 1)
we know that
The measure of the inner angle is the semi-sum of the arcs comprising it and its opposite.
we have
m∠1 -----> is the inner angle
The arcs that comprise it and its opposite are arc SP and arc QR
so
m∠1 =(1/2)[arc SP+arc QR]
Part 2)
we know that
The <u>Intersecting Secant-Tangent Theorem,</u> states that 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.
so
In this problem we have that
![PR^{2} =PS*PT](https://tex.z-dn.net/?f=PR%5E%7B2%7D%20%3DPS%2APT)
Part 3)
we know that
The <u>Tangent-Tangent Theorem</u> states that if from one external point, two tangents are drawn to a circle then they have equal tangent segments
so
In this problem
PQ=PR
Part 4)
we know that
The measurement of the outer angle is the semi-difference of the arcs it encompasses.
In this problem
m∠QPT -----> is the outer angle
The arcs that it encompasses are arc QT and arc QS
therefore
m∠QPT=(1/2)[arc QT-arc QS]
Answer:
An experimental study because it will give the most accurate results.
Step-by-step explanation:
The observational study could also be somewhat useful, but the most accurate results will come from an experimental study. You will test the difference in DNA, blood pressure, and such. Whilst, the observational study, you only can observe the effects from the outside. The sample studies mean nothing because you can't test both sides of the complete study.
Answer:
The answer is
![f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%7B%5Cdisplaystyle%20%208%20%2B%20%5Cfrac%7B-28%7D%7B1%7D%28x%2B2%29%2B%5Cfrac%7B-36%7D%7B2%21%7D%28x%2B2%29%5E2%20%2B%20%5Cfrac%7B-18%7D%7B3%21%7D%28x%2B2%29%5E2%20%7D)
Step-by-step explanation:
Remember that Taylor says that
![f(x) = {\displaystyle \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a) }{k!}(x-a)^k }](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%7B%5Cdisplaystyle%20%5Csum%5Climits_%7Bk%3D0%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7Bf%5E%7B%28k%29%7D%28a%29%20%7D%7Bk%21%7D%28x-a%29%5Ek%20%7D)
For this case
![f^{(0)} (-2) = 8(-2)-3(-2)^3 = 8\\f^{(1)} (-2) = 8-3(3)(-2)^2 = -28\\f^{(2)} (-2) = -3(3)2(-2) = -36\\f^{(2)} (-2) = -3(3)2 = -18](https://tex.z-dn.net/?f=f%5E%7B%280%29%7D%20%28-2%29%20%3D%208%28-2%29-3%28-2%29%5E3%20%3D%208%5C%5Cf%5E%7B%281%29%7D%20%28-2%29%20%3D%208-3%283%29%28-2%29%5E2%20%3D%20-28%5C%5Cf%5E%7B%282%29%7D%20%28-2%29%20%3D%20-3%283%292%28-2%29%20%3D%20-36%5C%5Cf%5E%7B%282%29%7D%20%28-2%29%20%3D%20-3%283%292%20%3D%20-18)
![f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%7B%5Cdisplaystyle%20%208%20%2B%20%5Cfrac%7B-28%7D%7B1%7D%28x%2B2%29%2B%5Cfrac%7B-36%7D%7B2%21%7D%28x%2B2%29%5E2%20%2B%20%5Cfrac%7B-18%7D%7B3%21%7D%28x%2B2%29%5E2%20%7D)