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3 years ago

Peer review involves scientists openly publishing details of the methods they used, the results of

Mathematics

1 answer:

IrinaK [193]3 years ago

4 0

Answer: 1. Peer review involves scientists openly publishing details of the methods they used, the results of

their experiments, and the reasoning behind their hypotheses for other scientists working in the

same field to evaluate. = B true .

2. Data from controlled experiments can provide evidence that one variable causes another. = B true .

Step-by-step explanation:

Peer review is when a scientist conducts an experiment and then gives all of their information and findings to peers in order for them to follow the findings and see if the first findings are rational.

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Can someone help me please

vitfil [10]

Answer:i would say the answer is A

Step-by-step explanation:its my personal opinion i hope it helps

6 0

2 years ago

Read 2 more answers

How do you find the property of 34 + 0 + 18 + 26 using Common Core math?

Flauer [41]

I assume you're looking for the sum.

I would use the null property to eliminate 0
34+0+18+26
=34+18+26
then use commutativity to rearrange
=34+26 + 18
Add 34+26=60
=60+18
=78

8 0

3 years ago

A parabola can be drawn given a focus of (8, 6) and a directrix of x=4. Write the equation of the parabola

Orlov [11]

Answer:

The equation of the Parabola

(y - 6 )² = 8 (x -6)

Step-by-step explanation:

Step(i):-

Given directrix x = 4

we know that x = h - a = 4

h -a = 4 ...(i)

Given Focus = ( 8,6)

we know that the Focus of the Parabola

( h + a , k ) = ( 8,6)

comparing h + a = 8 ...(ii)

k = 6

solving (i) and (ii) and adding

h - a + h+ a = 8 +4

2 h = 12

h =6

Put h = 6 in equation (i)

⇒ h - a =4

⇒ 6 - 4 = a

⇒ a = 2

Step(ii):-

The equation of the Parabola ( h,k) = (6 , 6)

( y - k )² = 4 a ( x - h )

(y - 6 )² = 4 (2) (x -6)

(y - 6 )² = 8 (x -6)

3 0

2 years ago

sarah is putting her sweaters in boxes to organize her closet.seven sweaters can fit in each box. sarah will fill as many boxes

Morgarella [4.7K]

Given: Sarah can fit 7 sweaters in a box

She has 38 sweaters

the remaining sweaters will stack on the shelf in the closet.

By dividing 38 by 7 we will get:

$\frac{38}{7}=\frac{35+3}{7}=\frac{5\cdot7+3}{7}=5\frac{3}{7}$

So, she will need 5 boxes, she will put 35 sweaters in 5 boxes

The remaining sweaters = 38 - 35 = 3 sweaters

So, the answer is: 3 sweaters will go in her closet

6 0

1 year ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

2 years ago