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1 year ago

5+2x=2x+6 What's the answer?

Mathematics

1 answer:

Marina86 [1]1 year ago

3 0

Answer:

Below!

Step-by-step explanation:

We can solve this with the help of equations.

5 + 2x = 2x + 6
=> -2x + 2x = -5 + 6
=> 0 ≠ 1

Since we are getting 0 ≠ 1 as our result, the problem has no solution.

Hoped this helped.

$BrainiacUser1357$

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How many ways can eight letters be arranged into groups of five where order matters and the first two letters are already chosen

fenix001 [56]

Answer:

120

Step-by-step explanation:

Since we're dealing with a problem where the order matters and the first two letters are already chosen we need to subtract the number of letters and the number of available slots per group.

We use the permutation formula to find the answer, but before that let's check values.

n = 8

k = 5

Now since there are two letters already chosen we have to deduct two from both the value of n and k.

n = 6

k = 3

Now we can use the permutation formula:

$_{n}P_{k}=\dfrac{n!}{(n-k)!}$

$_{6}P_{3}=\dfrac{6!}{6-3)!}$

$_{6}P_{3}=\dfrac{6!}{3!}$

$_{6}P_{3}=\dfrac{6*5*4*3*2*1}{3*2*1}$

The 3*2*1 cancels out and leaves us with:

$_{6}P_{3}=6*5*4$

$_{6}P_{3}=120$

So there are 120 possible ways to arrange eight letters into groups of five where order matters and the first two letters are already chosen.

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2 years ago

Read 2 more answers

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Mice21 [21]

The answer is 0.687.

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2 years ago

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Use the Midpoint Rule with n = 5 to estimate the volume V obtained by rotating about the y-axis the region under the curve y = 1

velikii [3]

Using the shell method, the volume is given exactly by the definite integral,

$2\pi\displaystyle\int_0^1x(1+9x^3)\,\mathrm dx$

Splitting up the interval [0, 1] into 5 subintervals gives the partition,

[0, 1/5], [1/5, 2/5], [2/5, 3/5], [3/5, 4/5], [4/5, 5]

with left and right endpoints, respectively, for the $i$ -th subinterval

$\ell_i=\dfrac{i-1}5$

$r_i=\dfrac i5$

where $1\le i\le5$ . The midpoint of each subinterval is

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}{10}$

Then the Riemann sum approximating the integral above is

$2\pi\displaystyle\sum_{i=1}^5m_i(1+9{m_i}^3)\frac{1-0}5$

$\dfrac{2\pi}5\displaystyle\sum_{i=1}^5\left(\frac{2i-1}{10}+9\left(\frac{2i-1}{10}\right)^4\right)$

$\dfrac{2\pi}{5\cdot10^4}\displaystyle\sum_{i=1}^5\left(16i^4-32i^3+24i^2+1992i-999\right)=\frac{112,021\pi}{25,000}\approx\boxed{14.08}$

(compare to the actual value of the integral of about 14.45)

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2 years ago

Factorise x 2 + 2 x − 48

Hatshy [7]

Answer:

x^2+ 2x +-48

product: -48 sum: 2

8, -6

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Answer:

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