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ladessa [460]
3 years ago
10

Help me please thank you

Mathematics
1 answer:
8090 [49]3 years ago
5 0

Answer:

D:(1,1)

Step-by-step explanation:

Make the ghraphic and you'll see

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damaskus [11]

Answer:

  1. Red: -2
  2. Blue: 5
  3. Yellow: -4
  4. Green: 3
  5. Orange: -1

5 0
3 years ago
Read 2 more answers
Which formula can be used to describe the sequence? –81, 108, –144, 192, ...
ale4655 [162]

Answer:

a(n) = -81(-4/3)^(n - 1)

Step-by-step explanation:

The first term is -81.  From -81 we obtain the next term, 108, by multiplying -81 by -4/3.  Note how (-4/3)(108) = -144 as anticipated.

Thus, the common ratio is -4/3.

The explicit function for this geometric series is

a(n) = -81(-4/3)^(n - 1)

8 0
3 years ago
Dilation of 4 n (0,0) k(1, 1) I (1, 0)
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5 0
3 years ago
Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n
Darina [25.2K]

Recall that for |<em>x</em>| < 1, we have

\displaystyle \frac1{1-x} = \sum_{n=0}^\infty x^n

Differentiating both sides gives

\displaystyle \frac1{(1-x)^2} = \sum_{n=0}^\infty nx^{n-1} = \sum_{n=0}^\infty (n+1)x^n

and multiplying both sides by 8 gives the series for <em>f(x)</em> :

f(x)=\displaystyle \frac8{(1-x)^2} = \boxed{8\sum_{n=0}^\infty (n+1)x^n}

and this converges over the same interval, |<em>x</em>| < 1, so that the radius of convergence is 1.

4 0
3 years ago
Help solve for “q”<br> —————————————
VMariaS [17]

Digram:-

\\

\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\put(5,1){\vector(1,0){4}}\put(5,1){\vector(-1,0){4}}\put(5,1){\vector(1,1){3}}\put(2,2){$\underline{\boxed{\large\sf a + b = 180^{\circ}}$}}\put(4.5,1.3){$\sf a^{\circ}$}\put(5.7,1.3){$\sf b^{\circ}$}\end{picture}

\\

STEP :-

\dashrightarrow \tt(4q - 1) {}^{ \circ}  +  {117}^{ \circ}  = 18 {0}^{ \circ}

{Linear pair}

\\  \\

\dashrightarrow \tt(4q - 1) {}^{ \circ}= 18 {0}^{ \circ}   - {117}^{ \circ}

\\

\dashrightarrow \tt(4q - 1) {}^{ \circ}=63^{ \circ}

\\

\dashrightarrow \tt4q - 1{}^{ \circ}=63^{ \circ}

\\

\dashrightarrow \tt4q =63^{ \circ} + 1{}^{ \circ}

\\

\dashrightarrow \tt4q =64{}^{ \circ}

\\

\dashrightarrow \tt \: q =  \dfrac{64}{4}^{ \circ}

\\

\dashrightarrow \tt \: q =  \dfrac{16 \times 4}{4}^{ \circ}

\\

\dashrightarrow \tt \: q =  \dfrac{16 \times \cancel4}{\cancel4}^{ \circ}

\\

\dashrightarrow \tt \: q =  \dfrac{16}{1}

\\

\dashrightarrow \bf q = 16 \degree

\\  \\

Verification:

\\

\dashrightarrow \tt(4 \times 16- 1) {}^{ \circ}  +  {117}^{ \circ}  = 18 {0}^{ \circ}

\\

\dashrightarrow \tt(64- 1) {}^{ \circ}  +  {117}^{ \circ}  = 18 {0}^{ \circ}

\\

\dashrightarrow \tt63^{ \circ}  +  {117}^{ \circ}  = 18 {0}^{ \circ}

\\

\dashrightarrow \tt180^{ \circ}  = 18 {0}^{ \circ}

\\

LHS = RHS

HENCE VERIFIED!

8 0
3 years ago
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