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Tpy6a [65]
3 years ago
11

If two lines intersect in one point, then exactly one plane contains both lines.

Mathematics
1 answer:
horsena [70]3 years ago
5 0
The answer is theorem.     
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Kaynard's first step in solving this rational equation is to simplify it by multiplying both sides by a common denominator. Whic
bulgar [2K]

x^2 - 9 = (x + 3)(x -3)

3x - 9 = 3(x - 3)

and

6

Common denominator would be 6(x+3)(x - 3)

Answer: 6(x+3)(x - 3)

5 0
3 years ago
F(x) =-2x^2-4x-18 find f (-6)
Alexxandr [17]

Answer:

ncdjndndd

Step-by-step explanation:

jdsjkdsjdsjkjkds

7 0
4 years ago
20 POINTS PLEAE HELP!!!!!!!!!!!!!!!
Alina [70]
His first payment is $100, thus a₁ = 100.

the next "term", month will be 1.1 times more than the one before, namely r = 1.1, the common ratio.

\bf \qquad \qquad \textit{sum of a finite geometric sequence}
\\\\
S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
a_1=100\\
r=1.1\\
n=20
\end{cases}

\bf \sum\limits_{i=1}^{20}~100(1.1)^{i-1}\qquad \qquad\qquad  \qquad S_{20}=100\left( \cfrac{1-1.1^{20}}{1-1.1} \right)
\\\\\\
S_{20}=100\left( \cfrac{1-\stackrel{\approx}{6.727499949}}{-0.1} \right)\implies S_{20}\approx 100(57.27499949)
\\\\\\
S_{20}\approx 5727.4999493256

is the serie divergent or convergent?

well, to make it short, when the common ratio is 0 < | r | < 1, namely a fraction between 0 and 1, only then the serie is convergent, namely it reaches a fixed value, now in this case, 1.1 is a value larger than anything between 0 and 1, so no dice.
5 0
3 years ago
Read 2 more answers
The scores on a test have a mean of 78 and a standard deviation of 6.5. A student scored a 67 on the test. What is the Z-Score a
Komok [63]

Answer:

76

Step-by-step explanation:

5 0
3 years ago
Use integration by parts to derive the following formula from the table of integrals.
emmasim [6.3K]

Answer:

I= (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C (for a≠0)

Step-by-step explanation:

for

I= ∫x^n . e^ax dx

then using integration by parts we can define u and dv such that

I= ∫(x^n) . (e^ax dx) = ∫u . dv

where

u= x^n → du = n*x^(n-1) dx

dv= e^ax  dx→ v = ∫e^ax dx = (e^ax) /a ( for a≠0 .when a=0 , v=∫1 dx= x)

then we know that

I= ∫u . dv = u*v - ∫v . du + C

( since d(u*v) = u*dv + v*du → u*dv = d(u*v) - v*du → ∫u*dv = ∫(d(u*v) - v*du) =

(u*v) - ∫v*du + C )

therefore

I= ∫u . dv = u*v - ∫v . du + C = (x^n)*(e^ax) /a - ∫ (e^ax) /a * n*x^(n-1) dx +C = = (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C

I= (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C (for a≠0)

5 0
3 years ago
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