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4 years ago

Which ordered pair is not a point on the graph of y = –5x + 2?

Mathematics

1 answer:

amid [387]4 years ago

6 0

Answer:

Option A. (-1, 6)

Step-by-step explanation:

If a point (x, y) lies on a graph, then the ordered pairs will satisfy the equation graphed.

For point (-1, 6)

y = -5x + 2

6 = (-5)(-1) + 2

6 = -5 + 2

6 = -3 [Not true]

Therefore, (-1, 6) doesn't lie on this graph.

For point (0, 2),

y = -5x + 2

2 = 0 + 2

2 = 2 [True]

Therefore, (0, 1) lies on the graph.

For point (-2, 12),

y = -5x + 2

12 = (-5)(-2) + 2

12 = 12 {True}

Therefore, point (-2, 12) lies on the graph.

For point (2, -8)

y = -5x + 2

-8 = (-5)(2) + 2

-8 = -8 [True}

Therefore, (2, -8) lies on the given graph

Option A. is the answer.

You might be interested in

Find the perimeter of △ABC with vertices A(−5, 3), B(4, −5), and C(−5, −5). Round your answer to the nearest hundredth.

sertanlavr [38]

The perimeter of △ABC with vertices A(−5, 3), B(4, −5), and C(−5, −5) is 29.04 units

<h3>Perimeter of a triangle</h3>

The perimeter of a triangle is the sum of the sides of the triangle.

Given the coordinate points A(−5, 3), B(4, −5), and C(−5, −5)

Determine the distance AB, AC and BC using the distance formula

AB = √(-5-3)²+(4+5)²

AB = √64+81
AB = √145

AB = 12.04 units

AC = √(-5-3)²+(-5+5)²

AC = √64

AC = 8 units

BC = √(-5-4)²+(-5+5)²

BC = √81
BC = 9 units

Take the sum

Perimeter of triangle ABC = 9 + 8 + 12.04

Perimeter of triangle ABC = 29.04 units

Hence the perimeter of △ABC with vertices A(−5, 3), B(4, −5), and C(−5, −5) is 29.04 units

Learn more on perimeter of triangle here: brainly.com/question/24382052

#SPJ1

8 0

2 years ago

Evaluate 1^3 + 2^3 +3^3 +.......+ n^3

Molodets [167]

Notice that

$(n+1)^4-n^4=4n^3+6n^2+4n+1$

so that

$\displaystyle\sum_{i=1}^n((n+1)^4-n^4)=\sum_{i=1}^n(4i^3+6i^2+4i+1)$

We have

$\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(2^4-1^4)+(3^4-2^4)+(4^4-3^4)+\cdots+((n+1)^4-n^4)$

$\implies\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(n+1)^4-1$

so that

$\displaystyle(n+1)^4-1=\sum_{i=1}^n(4i^3+6i^2+4i+1)$

You might already know that

$\displaystyle\sum_{i=1}^n1=n$

$\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2$

$\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$

so from these formulas we get

$\displaystyle(n+1)^4-1=4\sum_{i=1}^ni^3+n(n+1)(2n+1)+2n(n+1)+n$

$\implies\displaystyle\sum_{i=1}^ni^3=\frac{(n+1)^4-1-n(n+1)(2n+1)-2n(n+1)-n}4$

$\implies\boxed{\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4}$

If you don't know the formulas mentioned above:

The first one should be obvious; if you add $n$ copies of 1 together, you end up with $n$ .
The second one is easily derived: If $S=1+2+3+\cdots+n$ , then $S=n+(n-1)+(n-2)+\cdots+1$ , so that $2S=n(n+1)$ or $S=\dfrac{n(n+1)}2$ .
The third can be derived using a similar strategy to the one used here. Consider the expression $(n+1)^3-n^3=3n^2+3n+1$ , and so on.