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3 years ago

10

A fish tank has 10 goldfish, 6 tetras, 4 snails, and 2 platies. What is the ratio of tetras to platies?

1 answer:

Nat2105 [25]3 years ago

3 0

Answer:

6:2

Step-by-step explanation:

ummmmmm I'm done it's just that I can't do 20 words

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Find the sum of the first 47 terms of the following series, to the nearest integer.

Leokris [45]

Answer:

The sum of the first 47 terms of the series, 12, 16, 20, ... S₄₈ is 4,888

Step-by-step explanation:

The given series is;

12, 16, 20, ...,

Therefore, the first term of the series is, a = 12

The common difference of series is found as follows;

The difference between subsequent terms, 12 and 16 is 16 - 12 = 4

The difference between subsequent terms, 16, and 20 is 20 - 16 = 4

Therefore, the common difference, d = 4

The series is therefore an arithmetic projection, AP

The sum of the first 'n' terms of an AP, Sₙ, is given as follows;

$S_n = \dfrac{n}{2} \cdot \left [2 \cdot a + (n - 1)\cdot d \right ]$

(47/2)*(2*12+(47-1)*4)

The sum of the first 47 terms is therefore given as follows;

$S_n = \dfrac{47}{2} \cdot \left [2 \times 12 + (47 - 1)\times 4 \right ] = 4,888$

The sum of the first 47 terms of the series, 12, 16, 20, ... S₄₈ = 4,888

5 0

2 years ago

Solve the following equations: (a) x^11=13 mod 35 (b) x^5=3 mod 64

tino4ka555 [31]

a.

$x^{11}=13\pmod{35}\implies\begin{cases}x^{11}\equiv13\equiv3\pmod5\\x^{11}\equiv13\equiv6\pmod7\end{cases}$

By Fermat's little theorem, we have

$x^{11}\equiv (x^5)^2x\equiv x^3\equiv3\pmod5$

$x^{11}\equiv x^7x^4\equiv x^5\equiv6\pmod 7$

5 and 7 are both prime, so $\varphi(5)=4$ and $\varphi(7)=6$ . By Euler's theorem, we get

$x^4\equiv1\pmod5\implies x\equiv3^{-1}\equiv2\pmod5$

$x^6\equiv1\pmod7\impleis x\equiv6^{-1}\equiv6\pmod7$

Now we can use the Chinese remainder theorem to solve for $x$ . Start with

$x=2\cdot7+5\cdot6$

Taken mod 5, the second term vanishes and $14\equiv4\pmod5$ . Multiply by the inverse of 4 mod 5 (4), then by 2.

$x=2\cdot7\cdot4\cdot2+5\cdot6$

Taken mod 7, the first term vanishes and $30\equiv2\pmod7$ . Multiply by the inverse of 2 mod 7 (4), then by 6.

$x=2\cdot7\cdot4\cdot2+5\cdot6\cdot4\cdot6$

$\implies x\equiv832\pmod{5\cdot7}\implies\boxed{x\equiv27\pmod{35}}$

b.

$x^5\equiv3\pmod{64}$

We have $\varphi(64)=32$ , so by Euler's theorem,

$x^{32}\equiv1\pmod{64}$

Now, raising both sides of the original congruence to the power of 6 gives

$x^{30}\equiv3^6\equiv729\equiv25\pmod{64}$

Then multiplying both sides by $x^2$ gives

$x^{32}\equiv25x^2\equiv1\pmod{64}$

so that $x^2$ is the inverse of 25 mod 64. To find this inverse, solve for $y$ in $25y\equiv1\pmod{64}$ . Using the Euclidean algorithm, we have

64 = 2*25 + 14

25 = 1*14 + 11

14 = 1*11 + 3

11 = 3*3 + 2

3 = 1*2 + 1

=> 1 = 9*64 - 23*25

so that $(-23)\cdot25\equiv1\pmod{64}\implies y=25^{-1}\equiv-23\equiv41\pmod{64}$ .

So we know

$25x^2\equiv1\pmod{64}\implies x^2\equiv41\pmod{64}$

Squaring both sides of this gives

$x^4\equiv1681\equiv17\pmod{64}$

and multiplying both sides by $x$ tells us

$x^5\equiv17x\equiv3\pmod{64}$

Use the Euclidean algorithm to solve for $x$ .

64 = 3*17 + 13

17 = 1*13 + 4

13 = 3*4 + 1

=> 1 = 4*64 - 15*17

so that $(-15)\cdot17\equiv1\pmod{64}\implies17^{-1}\equiv-15\equiv49\pmod{64}$ , and so $x\equiv147\pmod{64}\implies\boxed{x\equiv19\pmod{64}}$

5 0

3 years ago

How do you do this? It's geometry

antiseptic1488 [7]

45.

12/18 = 8/x <=> 2/3 = 8/x <=> x = (3*8)/2 = 12;

7 0

3 years ago

Given g(x)=2x-1, solve for x when g(x)=3.

SashulF [63]

Answer:g(x)=2(3)+1=6+1=7

Step-by-step explanation:7

7 0

3 years ago

Read 2 more answers

Find f(5) if f(t) = ^6 – 3t – 4.<br> A –6 <br> B –2 <br> C 14 <br> D 6

Lera25 [3.4K]

What they said ^ I was just about to answer, but there’s popped up first lol

8 0

3 years ago