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marta [7]
3 years ago
11

Part A

Mathematics
1 answer:
Alekssandra [29.7K]3 years ago
6 0

Answer:

Step-by-step explanation:

There are several possible outcomes.  If your cross section is created by intersecting the cone with a plane parallel to the base of the cone, the cross section is a circle.  If the plane is held vertical, the cross section could be a hyperbola.  If the intersecting plane is neither vertical nor horizontal, the cross section could be an ellipse.  

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What are the potential solutions to the equation 2ln(x+3)=0
skad [1K]
2\ln(x+3)=0\implies \ln(x+3)^2=0\implies e^{\ln(x+3)^2}=e^0\implies (x+3)^2=1

Expanding the left side gives

x^2+6x+9=1\implies x^2+6x+8=0\implies (x+4)(x+2)=0

which gives two solutions, x=-4 and x=-2. But if x=-4, then \ln(x+3)=\ln(-1), but this number isn't real, so x=-4 is an extraneous solution. Meanwhile if x=-2, you get \ln(-2+3)=\ln1=0, so this solution is correct.

"Potential solutions" might refer to both possibilities, but there is only one actual (real) solution.
4 0
4 years ago
Read 2 more answers
The Bengals football team scored 2 touchdowns for 6 points each, one extra point, and 3 field goals for 3 points each. The Raven
Yanka [14]

bangles scored 22 pts but don't know how much the ravens scored


4 0
4 years ago
When people say 8:15 times 195 equals 15x, what does 15x mean or stand for
wel

It means that 15 times another number equals to your equation.


3 0
4 years ago
If someone around 831,982 to 830,000 and then someone else rounds 831,982 for 800,000 who would be correct
devlian [24]

<em><u></u></em>

<em><u></u></em>

Question

If someone around 831,982 to 830,000 and then someone else rounds 831,982 for 800,000 who would be correct?

Answer:

Hi, There! Mika-Chan I'm here to help! :)

<em><u>First We Need to know If we're rounding to the Nearest Hundred thousand or Not.</u></em>

<u><em>So The Answer would Be 800,000 If your Going to Round to the nearest Hundred </em></u>

<u><em>But If We're rounding to   The Nearest ten Thousand The Answer would be 830,000</em></u>

:D hope this Helps you!

3 0
3 years ago
Determine determine whether the following geometric series converges or diverges. if the series converges find its sum.
Lilit [14]

For starters,

\dfrac{3^k}{4^{k+2}}=\dfrac{3^k}{4^24^k}=\dfrac1{16}\left(\dfrac34\right)^k

Consider the nth partial sum, denoted by S_n:

S_n=\dfrac1{16}\left(\dfrac34\right)+\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\cdots+\dfrac1{16}\left(\dfrac34\right)^n

Multiply both sides by \frac34:

\dfrac34S_n=\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\dfrac1{16}\left(\dfrac34\right)^4+\cdots+\dfrac1{16}\left(\dfrac34\right)^{n+1}

Subtract S_n from this:

\dfrac34S_n-S_n=\dfrac1{16}\left(\dfrac34\right)^{n+1}-\dfrac1{16}\left(\dfrac34\right)

Solve for S_n:

-\dfrac14S_n=\dfrac3{64}\left(\left(\dfrac34\right)^n-1\right)

S_n=\dfrac3{16}\left(1-\left(\dfrac34\right)^n\right)

Now as n\to\infty, the exponential term will converge to 0, since r^n\to0 if 0. This leaves us with

\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=1}^n\frac{3^k}{4^{k+2}}=\sum_{k=1}^\infty\frac{3^k}{4^{k+2}}=\frac3{16}

8 0
3 years ago
Read 2 more answers
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