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rosijanka [135]
3 years ago
14

If two lines have the same slope, they are _______.

Mathematics
1 answer:
Sever21 [200]3 years ago
4 0

Answer:

same slope is parallel lines

Step-by-step explanation:

You might be interested in
Write the decimal equivalent for this fraction 70/100
d1i1m1o1n [39]

Answer:

0.7 is the decimal equivalent to 70/100.

Step-by-step explanation:

You can just do the numerator (70) divided by the denominator (100): 70÷100 = 0.7

8 0
2 years ago
56% of 860 can anyone help please?
Tems11 [23]
The answer is 481.6, because 56% = 0.56, so just multiply 860 by 0.56 and u get 481.6
4 0
3 years ago
Read 2 more answers
2a is any number a multiplied by the number 2. Calculate the value of 2a for a = 7.
marshall27 [118]

Answer:

14

Step-by-step explanation:

step 1.

substitute 'a' with 7

You are then left with a multiplication problem.

step 2.

solve 2 × 7

answer is 14

3 0
3 years ago
25 markers in a bag: 5 red, 5 yellow, 5 blue, 5 green, 5 purple One marker is drawn out of the bag, then put back in. How many t
lawyer [7]

Answer: 16

Step-by-step explanation:

From the question, we are informed that there are 25 markers in a bag: 5 red, 5 yellow, 5 blue, 5 green, 5 purple and that one marker is drawn out of the bag, then put back in.

The probability of picking a blue marker will be: = 5/25 = 1/5

Therefore, the number of times that a blue marker will be expected to be picked in 80 draws would be:

= 1/5 × 80

= 16 times

7 0
3 years ago
Read 2 more answers
Solve for y where y(2)=2 and y'(2)=0 by representing y as a power series centered at x=a
Crank

I'll assume the ODE is actually

y''+(x-2)y'+y=0

Look for a series solution centered at x=2, with

y=\displaystyle\sum_{n\ge0}c_n(x-2)^n

\implies y'=\displaystyle\sum_{n\ge0}(n+1)c_{n+1}(x-2)^n

\implies y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n

with c_0=y(2)=2 and c_1=y'(2)=0.

Substituting the series into the ODE gives

\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge0}(n+1)c_{n+1}(x-2)^{n+1}+\sum_{n\ge0}c_n(x-2)^n=0

\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge0}c_n(x-2)^n=0

\displaystyle2c_2+c_0+\sum_{n\ge1}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge1}c_n(x-2)^n=0

\displaystyle2c_2+c_0+\sum_{n\ge1}\bigg((n+2)(n+1)c_{n+2}+(n+1)c_n\bigg)(x-2)^n=0

\implies\begin{cases}c_0=2\\c_1=0\\(n+2)c_{n+2}+c_n=0&\text{for }n>0\end{cases}

  • If n=2k for integers k\ge0, then

k=0\implies n=0\implies c_0=c_0

k=1\implies n=2\implies c_2=-\dfrac{c_0}2=(-1)^1\dfrac{c_0}{2^1(1)}

k=2\implies n=4\implies c_4=-\dfrac{c_2}4=(-1)^2\dfrac{c_0}{2^2(2\cdot1)}

k=3\implies n=6\implies c_6=-\dfrac{c_4}6=(-1)^3\dfrac{c_0}{2^3(3\cdot2\cdot1)}

and so on, with

c_{2k}=(-1)^k\dfrac{c_0}{2^kk!}

  • If n=2k+1, we have c_{2k+1}=0 for all k\ge0 because c_1=0 causes every odd-indexed coefficient to vanish.

So we have

y(x)=\displaystyle\sum_{k\ge0}c_{2k}(x-2)^{2k}=\sum_{k\ge0}(-1)^k\frac{(x-2)^{2k}}{2^{k-1}k!}

Recall that

e^x=\displaystyle\sum_{n\ge0}\frac{x^k}{k!}

The solution we found can then be written as

y(x)=\displaystyle2\sum_{k\ge0}\frac1{k!}\left(-\frac{(x-2)^2}2\right)^k

\implies\boxed{y(x)=2e^{-(x-2)^2/2}}

6 0
3 years ago
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