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yuradex [85]
4 years ago
13

3x - 4y = 19; 8x + y = 12 Are the equations parallel, perpendicular, or neither

Mathematics
1 answer:
stellarik [79]4 years ago
8 0

Step-by-step explanation:

Hey, there!

Let's check whether the lines are parallel, perpendicular or neither.

Fistly let's check of parallel ,

Let me tell you when two st. lines are paralle, then their slope are equal.

Given equation are,

3x - 4y = 9.........(i)

8x+y = 12 ..........(ii)

Now,

From equation (i)

slope \: (m1) =   \frac{ - coffe. \: ofx}{coffe. \: of \: y}

\: slope \: (m1) =   \frac{ - 3}{ - 4}

therefore \: slope \: (m1) =  \frac{3}{4}

now, again slope from equation (ii).

slope(m2) =  \frac{ - coffe.of \: x}{coffe. \: of \: y}

slope \: (m2) =  \frac{ - 8}{1}

Therefore, the slope of equation (ii) is -8.

Since, Their slopes are not equal, they are not parallel.

Now, let's check for perpendicular,

To be perpendicular, slope (m1)× slope (m2)= -1

now,

=  \frac{3}{4} \times  - 8

= 3×-2

= -6.

So, the equations are neither parallel nor perpendicular.

<em>Hope</em><em> </em><em>it helps</em><em>.</em><em>.</em><em>.</em>

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Murljashka [212]

Answer:

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3 years ago
I need help with 2,3 and 6! Please Help!
Georgia [21]
2. If you already know Faulhaber's formula, which says

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

then it's just a matter of setting n=4. If you don't, then you can prove that it works (via induction), or compute the sum by some other means. Presumably you're not expected to use brute force and just add the squares of 1 through 4.

Just to demonstrate one possible method of verifying the formula, suppose we start from the binomial expansion of (k-1)^3, do some manipulation, then sum over 1\le k\le n:

(k-1)^3=k^3-3k^2+3k-1
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\implies\displaystyle\sum_{k=1}^n(k^3-(k-1)^3)=\sum_{k=1}^n(3k^2-3k+1)

The left side is a telescoping series - several terms in consecutive terms of the series will cancel - and reduces to n^3. For example,

\displaystyle\sum_{k=3}^2(k^3-(k-1)^3)=(1^3-0^3)+(2^3-1^3)+(3^3-2^3)=3^3

Distributing the sum on the right side across each term and pull out constant factors to get

\displaystyle n^3=3\sum_{k=1}^nk^2-3\sum_{k=1}^nk+\sum_{k=1}^n1

If you don't know the formula for \displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2, you can use a similar trick with the binomial expansion (k-1)^2, or a simpler trick due to Gauss, or other methods. I'll assume you know it to save space for the other parts of your question. We then have

\displaystyle n^3=3\sum_{k=1}^nk^2-\frac{3n(n+1)}2+n
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and when n=4 we get 30.

3. Each term in the sum is a cube, but the sign changes. Recall that (-1)^n is either 1 if n is even or -1 if n is odd. So we can write

1^3-2^3+3^3-4^3+5^3=\displaystyle\sum_{k=1}^5(-1)^{k-1}k^3

(k+1 as the exponent to -1 also works)

6. If 0\le k\le n, and i=k+1, then we would get 1\le k+1\le n+1\iff1\le i\le n+1. So the sum with respect to i is

\displaystyle\sum_{k=0}^n\frac{k^2}{k+n}=\sum_{i=1}^{n+1}\frac{(i-1)^2}{i+n-1}
4 0
3 years ago
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Answer:

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4 0
3 years ago
(1.3t3 + 0.4t2 – 24t) – (0.6t2 + 8 – 18t)
lorasvet [3.4K]
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leva [86]

Answer:

The proportion of trees greater than 5 inches is expected to be 0.25 of the total amount of trees.

Step-by-step explanation:

In this problem we have a normal ditribution with mean of 4.0 in and standard deviation of 1.5 in.

The proportion of the trees that are expected to have diameters greater than 5 inches is equal to the probability of having a tree greater than 5 inches.

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