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Elanso [62]
3 years ago
5

25 points!!! Find first differences for the sequence in order from a1 to a5. Determine whether or not the series is quadratic or

not.
-1, -3, -1, 5, 15

a) 2, 2, 6, 10; not quadratic
b) 2, 2, 6, 10; quadratic
c) -2, 2, 6, 10; not quadratic
d) -2, 2, 6, 10; quadratic
Mathematics
1 answer:
crimeas [40]3 years ago
8 0
The answer would best be b hope this helps
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Can someone please help me with this
oee [108]

Answer:

165 pennies

Step-by-step explanation:

250/1.52=164.473684211, but just round it to the nearest whole number whitch would be 165!

6 0
2 years ago
Mervin had some cartons of milk he sold 2/5 of the cartons of milk in the morning he then sold 2/4 of the remainder in the morni
agasfer [191]
4/5 mervin had at first


3 0
3 years ago
Choose the correct word that completes each statement about inscribing a square in a circle.
Anna11 [10]

Answer:

1. Perpendicular

2. Isosceles

3. Never

Step-by-step explanation:

1. AC ⊥ BD because diameter of a square are perpendicular bisector of each other.

2. In Δ AOB , By using pythagoras : AB² = OA² + OB² .......( 1 )

In Δ COB , By using pythagoras : BC² = OC² + OB²  ..........( 2 )

But, OA = OC because both are radius of same circle

So, by using equations ( 1 ) and ( 2 ), We get AB = BC ≠ AC

⇒ ABC is a triangle having two equal sides so ABC is an isosceles triangle.

3. The side can never be equal to radius of circle because the side of the square will be chord for the circle and in a circle chord can never be equal to its radius


7 0
3 years ago
Graph the circle x^2+y^2+2x+4y-44=0
Natasha_Volkova [10]

Answer:

See explanation

Step-by-step explanation:

First, convert this to standard form. By completing the square, you can do the following:

(x^2+2x+1)-1+(y^2+4y+4)-4-44=0

(x+1)^2+(y+2)^2=49

This is a circle with center (-1,-2), and radius 7. It looks like the one I graphed below. Hope this helps!

6 0
3 years ago
The sum of three numbers in <br> g.p. is 21 and the sum of their squares is 189. find the numbers.
sashaice [31]
Let the three gp be a, ar and ar^2
a + ar + ar^2 = 21 => a(1 + r + r^2) = 21 . . . (1)
a^2 + a^2r^2 + a^2r^4 = 189 => a^2(1 + r^2 + r^4) = 189 . . . (2)
squaring (1) gives
a^2(1 + r + r^2)^2 = 441 . . . (3)
(3) ÷ (2) => (1 + r + r^2)^2 / (1 + r^2 + r^4) = 441/189 = 7/3
3(1 + r + r^2)^2 = 7(1 + r^2 + r^4)
3(r^4 + 2r^3 + 3r^2 + 2r + 1) = 7(1 + r^2 + r^4)
3r^4 + 6r^3 + 9r^2 + 6r + 3 = 7 + 7r^2 + 7r^4
4r^4 - 6r^3 - 2r^2 - 6r + 4 = 0
r = 1/2 or r = 2
From (1), a = 21/(1 + r + r^2)
When r = 2:
a = 21/(1 + 2 + 4) = 21/7 = 3
Therefore, the numbers are 3, 6 and 12.
3 0
3 years ago
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