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2 years ago

The given equations below are parallel perpendicular or neither

Mathematics

2 answers:

Ksju [112]2 years ago

5 0

The given equations below are parallel, because they have the same slope.

We have a general equation:

y=mx+b

m=slope.

In this case the slope of y=-1/3 x+1 will be m=-1/3

and the slope of y=-1/3 x+3 will be m=-1/3

Because they have the same slope they will be parallel.

pentagon [3]2 years ago

3 0

Answer:

Parallel

Step-by-step explanation:

Both slopes are the same. m = m

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1 year ago

Solve for the roots in simplest form using the quadratic formula: 4x2 + 4x = -5

weqwewe [10]

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Step-by-step explanation:

Move all terms to one side

4x^2 + 4x + 5 = 0

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2 years ago

Suppose you choose a team of two people from a group of n > 1 people, and your opponent does the same (your choices are allow

jonny [76]

Answer:

The number of possible choices of my team and the opponents team is

$\left\begin{array}{ccc}n-1\\E\\n=1\end{array}\right i^{3}$

Step-by-step explanation:

selecting the first team from n people we have $\left(\begin{array}{ccc}n\\1\\\end{array}\right) = n$ possibility and choosing second team from the rest of n-1 people we have $\left(\begin{array}{ccc}n-1\\1\\\end{array}\right) = n-1$

As { A, B} = {B , A}

Therefore, the total possibility is $\frac{n(n-1)}{2}$

Since our choices are allowed to overlap, the second team is $\frac{n(n-1)}{2}$

Possibility of choosing both teams will be

$\frac{n(n-1)}{2} * \frac{n(n-1)}{2} \\\\= [\frac{n(n-1)}{2}] ^{2}$

We now have the formula

1³ + 2³ + ........... + n³ = $[\frac{n(n+1)}{2}] ^{2}$

1³ + 2³ + ............ + (n-1)³ = $[x^{2} \frac{n(n-1)}{2}] ^{2}$

= $\left[\begin{array}{ccc}n-1\\E\\i=1\end{array}\right] = [\frac{n(n-1)}{2}]^{3}$

4 0

3 years ago

The given equations below are parallel perpendicular or neither​

The given equations below are parallel perpendicular or neither