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3 years ago

What is the 250th term of this sequence. Please explain. 4, 9,14,19,24

1 answer:

4 0

A1 = first term = 4
a2 = second term = a1+5 = 4+5 = 0
a3 = third term a1 + 5 + 5 = a1 + 5(n-1), where n is the subscript that represents which term we're discussing.

an=4+5(n-1).

Is this correct? Let's check it and find out.
What is your prediction for a2? Here, n = 2. Then a2 = 4+5(2-1), or 4+5, or 9. That agrees with the given sequence rule.

Thus, the 250th term would be 4+5(250-1). Evaluate this, please.

You might be interested in

Which of the following is not a Pythagorean triple?

Ierofanga [76]

Answer:

(7, 24, 26)

Step-by-step explanation:

A Pythagorean triple must have an odd number of even numbers. The triple (7, 24, 26) is not a Pythagorean triple.

_____

Additional comment

For an odd integer n, a triple can be formed as ...

(n, (n²-1)/2, (n²+1)/2)

That is, the following will be Pythagorean triples.

(3, 4, 5)
(5, 12, 13)
(7, 24, 25)
(9, 40, 41)
(11, 60, 61)

Another series involves even numbers and numbers separated by 2:

(2n, n²-1, n²+1)

(8, 15, 17)
(12, 35, 37)
(16, 63, 65)

In this list, if n is not a multiple of 2, the triple will be a multiple of one from the odd-number series.

It is a good idea to remember a few of these, as they tend to show up in Algebra, Geometry, and Trigonometry problems.

8 0

2 years ago

Ivan used coordinate geometry to prove that quadrilateral EFGH is a square.

Gelneren [198K]

Answer:

(A)Segment EF, segment FG, segment GH, and segment EH are congruent

Step-by-step explanation:

Step 1

Quadrilateral EFGH with points E(-2,3), F(1,6), G(4,3), H(1,0)

Step 2

Using the distance formula

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Given E(-2,3), F(1,6)

$|EF|=\sqrt{(6-3)^2+(1-(-2))^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}$

Given F(1,6), G(4,3)

$|FG|=\sqrt{(3-6)^2+(4-1)^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}$

Given G(4,3), H(1,0)

$|GH|=\sqrt{(0-3)^2+(1-4)^2}=\sqrt{(-3)^2+(-3)^2}=\sqrt{18}=3\sqrt{2}$

Given E (−2, 3), H (1, 0)

$|EH|=\sqrt{(0-3)^2+(1-(-2))^2}=\sqrt{(-3)^2+(3)^2}=\sqrt{18}=3\sqrt{2}$

Step 3

Segment EF ,E (−2, 3), F (1, 6)

Slope of |EF|= $\frac{6-3}{1+2} =\frac{3}{3}=1$

Segment GH, G (4, 3), H (1, 0)

Slope of |GH| $= \frac{0-3}{1-4} =\frac{-3}{-3}=1$

Step 4

Segment EH, E(−2, 3), H (1, 0)

Slope of |EH| $= \frac{0-3}{1+2} =\frac{-3}{3}=-1$

Segment FG, F (1, 6,) G (4, 3)

Slope of |EH| $=\frac{3-6}{4-1} =\frac{-3}{3}=-1$

Step 5

Segment EF and segment GH are perpendicular to segment FG.

The slope of segment EF and segment GH is 1. The slope of segment FG is −1.

Step 6

Segment EF, segment FG, segment GH, and segment EH are congruent.

The slope of segment FG and segment EH is −1. The slope of segment GH is 1.

Step 7

All sides are congruent, opposite sides are parallel, and adjacent sides are perpendicular. Quadrilateral EFGH is a square

4 0

3 years ago

Read 2 more answers

Help me please help

solniwko [45]

Answer:

I think its c I'm not sure

4 0

3 years ago

Please help me with algebra part B and C. attachment below. Will mark as brainslist. 25 points.

miv72 [106K]

Answer:

C. $(-7x+60)(x+4)$

Step-by-step explanation:

Consider the expression $-7x^2+32x+240.$

First, note that

$a=-7\\ \\b=32\\ \\c=240$

Find the discriminant

$D=b^2-4ac=32^2-4\cdot (-7)\cdot 240=1,024+6,720=7,744\\ \\\sqrt{D}=\sqrt{7,744}=88$

Now,

$x_1=\dfrac{-b+\sqrt{D}}{2a}=\dfrac{-32+88}{2\cdot (-7)}=-4\\ \\x_2=\dfrac{-b-\sqrt{D}}{2a}=\dfrac{-32-88}{2\cdot (-7)}=\dfrac{60}{7}$

Write the factored form:

$-7x^2+32x+240=-7(x-(-4))\left(x-\dfrac{60}{7}\right)=(x+4)(-7x+60)$

8 0

3 years ago

Diego solved an equation by multiplying both sides of the equation by 6. Then he checked that 6 is the correct solution by subst

Maslowich

I believe your answer would be c

8 0

3 years ago

Read 2 more answers