The value of the given integral is (3x+2)^2^1 /63 +c
= ![1/3\int\limits^a_b {(3x-2)^2^0} \, dx](https://tex.z-dn.net/?f=1%2F3%5Cint%5Climits%5Ea_b%20%7B%283x-2%29%5E2%5E0%7D%20%5C%2C%20dx)
=1/3*{(3x-20)^21}/21 +c
=1/63 *(3x-21)^21 +c
=d/dx {1/3*(3x-2)^21}/21 +c
=3/3 *{21(3x-2)}/21 +c
Let u=3x-2
du= 3dx
=![\int\limits^a_b {(3x-2)^2^0} \, dx = 1/3\int\limits^a_b {(3x-2)^2^0 dx} \, 3dx](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%283x-2%29%5E2%5E0%7D%20%5C%2C%20dx%20%3D%201%2F3%5Cint%5Climits%5Ea_b%20%7B%283x-2%29%5E2%5E0%20dx%7D%20%5C%2C%203dx)
=1/3![\int\limits^a_b {u^2^0} \, du =1/63 * u^2^1](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7Bu%5E2%5E0%7D%20%5C%2C%20du%20%20%3D1%2F63%20%2A%20u%5E2%5E1)
=(3x+2)^2^1 /63 +c
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30.05 times 10= 300.5 you just move the decimal point further back
Answer:
1. First method : Since, ![1\% = \frac{1}{100}](https://tex.z-dn.net/?f=1%5C%25%20%3D%20%5Cfrac%7B1%7D%7B100%7D)
⇒ ![40\% = \frac{40}{100}](https://tex.z-dn.net/?f=40%5C%25%20%3D%20%5Cfrac%7B40%7D%7B100%7D)
Thus, 40 % of 60
= ![\frac{40}{100}\times 60 = \frac{2400}{100} = 24](https://tex.z-dn.net/?f=%5Cfrac%7B40%7D%7B100%7D%5Ctimes%2060%20%3D%20%5Cfrac%7B2400%7D%7B100%7D%20%3D%2024)
Second method: (Pictorial model)
When we want get any % of 60,
Then, ![100\% = 60](https://tex.z-dn.net/?f=100%5C%25%20%3D%2060)
⇒
⇒ ![40\% = 4\times 6 = 24](https://tex.z-dn.net/?f=40%5C%25%20%3D%204%5Ctimes%206%20%3D%2024)
2. First method :
Let the number is x,
Since, ![1\% = \frac{1}{100}](https://tex.z-dn.net/?f=1%5C%25%20%3D%20%5Cfrac%7B1%7D%7B100%7D)
⇒ ![15\% = \frac{15}{100}](https://tex.z-dn.net/?f=15%5C%25%20%3D%20%5Cfrac%7B15%7D%7B100%7D)
Thus, 15 % of x = 30
![\frac{15}{100}\times x = 30](https://tex.z-dn.net/?f=%5Cfrac%7B15%7D%7B100%7D%5Ctimes%20x%20%3D%2030)
![\frac{15x}{100}=30](https://tex.z-dn.net/?f=%5Cfrac%7B15x%7D%7B100%7D%3D30)
![15 x = 3000\implies x=200](https://tex.z-dn.net/?f=15%20x%20%3D%203000%5Cimplies%20x%3D200)
Second method: (Pictorial model)
![15\% = 30](https://tex.z-dn.net/?f=15%5C%25%20%3D%2030)
![1\% = \frac{30}{15} = 2](https://tex.z-dn.net/?f=1%5C%25%20%3D%20%5Cfrac%7B30%7D%7B15%7D%20%3D%202)
![100\%=200](https://tex.z-dn.net/?f=100%5C%25%3D200)
Therefore, answer is 200.
no fancy math here, just some reasoning with words. I just feel like it
for one step in x-direction the value of the function doubles, so the common factor is 2.
so let's trace it back
(1,-6)
(0, -3)
For an exponential function the y-intercept is the "initial value".
do let's write is as a function
![y = - 3 \times {2}^{x}](https://tex.z-dn.net/?f=y%20%3D%20%20%20-%203%20%5Ctimes%20%20%7B2%7D%5E%7Bx%7D%20)
you could also use the points and plug their numbers into the general formula, but I felt this explanation could give you more understanding, since it's meant to be more intuitive.
hope it helps
Answer:
Choose one:
ray LG
ray LI
ray LJ
Step-by-step explanation:
There are three possible correct answers in this problem.
You only need one of them, but I will give you all three, and you choose the one you prefer.
Answer 1.
Ray LG is an angle bisector.
m<KLH = 120°
m<KLG + m<GLH = m<KLH
60° + m<GLH = 120°
m<GLH = 60°
Since m<KLG + m<GLH = m<KLH, and m<GLH = m<KLG, then
ray LG is the angle bisector of <KLH.
Answer: ray LG
Answer 2.
Angles KLH and HLM are a linear pair and supplementary. Their measures add to 180°. Since m<KLH = 120°, then m<HLM = 60°.
m<HLI = 30°, so m<ILM = 30°.
That makes ray LI a bisector of angle HLM.
Answer: ray LI
Answer 3.
Since ray LI bisects <HLM, and m<HLM = 60, then m<ILM = 30°.
We are given m<JLM = 15°, so m<ILJ is also 15°. That makes ray LJ an angle bisector.
Answer: ray LJ