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Simora [160]
3 years ago
7

Solve the exponential equation for x 64^7x-8 = 1

Mathematics
2 answers:
SOVA2 [1]3 years ago
7 0

X=9/64^7

Hope this helps

sashaice [31]3 years ago
6 0

Answer:

2.4 x 10^-12

Step-by-step explanation:

that is the explanation for the question

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A 3-gallon jug of water costs $21.12. What is the price per cup?
Rainbow [258]

Answer:

Its 0.44

Step-by-step explanation:

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7 0
4 years ago
If cos θ= 12 /13 and θ is located in the Quadrant I, find sin (2 θ ), cos(2 θ ), tan(2 θ )
PIT_PIT [208]

Answer:

\cos 2 \theta = \dfrac{119}{169}\\\\\sin2 \theta  = \dfrac{120}{169}\\\\\tan 2\theta = \dfrac{120}{119}

Step-by-step explanation:

\text{Given that,} \cos \theta = \dfrac{12}{13}\\ \\\text{Now,}\\\\\cos 2 \theta = 2 \cos^2 \theta - 1\\\\\\~~~~~~~~~=2 \left( \dfrac{12}{13} \right)^2 - 1\\\\\\~~~~~~~~~=2 \left( \dfrac{144}{169} \right) - 1\\\\\\~~~~~~~~~=\dfrac{288}{169}-1\\\\\\~~~~~~~~~=\dfrac{119}{169}

\sin 2\theta = 2 \sin \theta \cos \theta\\\\\\~~~~~~~~=2\sqrt{1 -\cos^2 \theta} \cdot \cos \theta~~~~~~~~~~~;[\text{In quadrant I, all ratios are positive.}]\\\\\\~~~~~~~~=2 \sqrt{1- \left( \dfrac{12}{13} \right)^2} \cdot \left(\dfrac{12}{13} \right)\\\\\\~~~~~~~~=\left( \dfrac{24}{13} \right) \sqrt{1- \dfrac{144}{169}}\\\\\\~~~~~~~~=\dfrac{24}{13}\sqrt{\dfrac{25}{169}}\\\\\\~~~~~~~=\dfrac{24}{13} \times \dfrac{5}{13}\\\\\\~~~~~~=\dfrac{120}{169}

\tan 2 \theta = \dfrac{\sin 2\theta}{\cos 2\theta}\\\\\\~~~~~~~~~=\dfrac{\tfrac{120}{169}}{ \tfrac{119}{169}}\\\\\\~~~~~~~~~=\dfrac{120}{169} \times \dfrac{169}{119}\\\\\\~~~~~~~~~=\dfrac{120}{119}

5 0
2 years ago
Can someone help me please?
Helga [31]
You just find the lowest denominators
7 0
4 years ago
2,17,82,257,626,1297 next one please ?​
In-s [12.5K]

The easy thing to do is notice that 1^4 = 1, 2^4 = 16, 3^4 = 81, and so on, so the sequence follows the rule n^4+1. The next number would then be fourth power of 7 plus 1, or 2402.

And the harder way: Denote the <em>n</em>-th term in this sequence by a_n, and denote the given sequence by \{a_n\}_{n\ge1}.

Let b_n denote the <em>n</em>-th term in the sequence of forward differences of \{a_n\}, defined by

b_n=a_{n+1}-a_n

for <em>n</em> ≥ 1. That is, \{b_n\} is the sequence with

b_1=a_2-a_1=17-2=15

b_2=a_3-a_2=82-17=65

b_3=a_4-a_3=175

b_4=a_5-a_4=369

b_5=a_6-a_5=671

and so on.

Next, let c_n denote the <em>n</em>-th term of the differences of \{b_n\}, i.e. for <em>n</em> ≥ 1,

c_n=b_{n+1}-b_n

so that

c_1=b_2-b_1=65-15=50

c_2=110

c_3=194

c_4=302

etc.

Again: let d_n denote the <em>n</em>-th difference of \{c_n\}:

d_n=c_{n+1}-c_n

d_1=c_2-c_1=60

d_2=84

d_3=108

etc.

One more time: let e_n denote the <em>n</em>-th difference of \{d_n\}:

e_n=d_{n+1}-d_n

e_1=d_2-d_1=24

e_2=24

etc.

The fact that these last differences are constant is a good sign that e_n=24 for all <em>n</em> ≥ 1. Assuming this, we would see that \{d_n\} is an arithmetic sequence given recursively by

\begin{cases}d_1=60\\d_{n+1}=d_n+24&\text{for }n>1\end{cases}

and we can easily find the explicit rule:

d_2=d_1+24

d_3=d_2+24=d_1+24\cdot2

d_4=d_3+24=d_1+24\cdot3

and so on, up to

d_n=d_1+24(n-1)

d_n=24n+36

Use the same strategy to find a closed form for \{c_n\}, then for \{b_n\}, and finally \{a_n\}.

\begin{cases}c_1=50\\c_{n+1}=c_n+24n+36&\text{for }n>1\end{cases}

c_2=c_1+24\cdot1+36

c_3=c_2+24\cdot2+36=c_1+24(1+2)+36\cdot2

c_4=c_3+24\cdot3+36=c_1+24(1+2+3)+36\cdot3

and so on, up to

c_n=c_1+24(1+2+3+\cdots+(n-1))+36(n-1)

Recall the formula for the sum of consecutive integers:

1+2+3+\cdots+n=\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2

\implies c_n=c_1+\dfrac{24(n-1)n}2+36(n-1)

\implies c_n=12n^2+24n+14

\begin{cases}b_1=15\\b_{n+1}=b_n+12n^2+24n+14&\text{for }n>1\end{cases}

b_2=b_1+12\cdot1^2+24\cdot1+14

b_3=b_2+12\cdot2^2+24\cdot2+14=b_1+12(1^2+2^2)+24(1+2)+14\cdot2

b_4=b_3+12\cdot3^2+24\cdot3+14=b_1+12(1^2+2^2+3^2)+24(1+2+3)+14\cdot3

and so on, up to

b_n=b_1+12(1^2+2^2+3^2+\cdots+(n-1)^2)+24(1+2+3+\cdots+(n-1))+14(n-1)

Recall the formula for the sum of squares of consecutive integers:

1^2+2^2+3^2+\cdots+n^2=\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6

\implies b_n=15+\dfrac{12(n-1)n(2(n-1)+1)}6+\dfrac{24(n-1)n}2+14(n-1)

\implies b_n=4n^3+6n^2+4n+1

\begin{cases}a_1=2\\a_{n+1}=a_n+4n^3+6n^2+4n+1&\text{for }n>1\end{cases}

a_2=a_1+4\cdot1^3+6\cdot1^2+4\cdot1+1

a_3=a_2+4(1^3+2^3)+6(1^2+2^2)+4(1+2)+1\cdot2

a_4=a_3+4(1^3+2^3+3^3)+6(1^2+2^2+3^2)+4(1+2+3)+1\cdot3

\implies a_n=a_1+4\displaystyle\sum_{k=1}^3k^3+6\sum_{k=1}^3k^2+4\sum_{k=1}^3k+\sum_{k=1}^{n-1}1

\displaystyle\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}4

\implies a_n=2+\dfrac{4(n-1)^2n^2}4+\dfrac{6(n-1)n(2n)}6+\dfrac{4(n-1)n}2+(n-1)

\implies a_n=n^4+1

4 0
3 years ago
Please answer the question above and pick an answer from the 3 provided
shutvik [7]

Answer:

B.

w=P-2l/2

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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