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3 years ago

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Write an equation in which the quadratic expression 3x^2+6x-24 equals 0. Show the expression in factored form and explain what y

our solutions mean for the equation. Show your work.

1 answer:

mr_godi [17]3 years ago

5 0

Answer:

3x² + 6x - 24 = 0

(3x - 6)(x + 4) = 0

x = -4 , x = 2 when y = 0

Step-by-step explanation:

3x² + 6x - 24 = 0

(3x - 6)(x + 4) = 0

Means

3x - 6 = 0 ⇒ 3x = 6 ⇒ x = 6 ÷ 3 = 2

x + 4 = 0 ⇒ x = -4

The Parabola which represents the quadratic equation intersects x-axis at

points (2 , 0) and (-4 , 0)

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H (x)=x^2+ 8x+7 , what function family does this equation belong to?

Semenov [28]

Answer:

Algebra

Step-by-step explanation:

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3 years ago

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Can you guys distribute -9n-8-3n=6n-8 and figure out how many solutions it has plz

nydimaria [60]

First combine like terms (-9n-3n=-12n)
Then, subtract 6n on both sides (to get rid of the 6n) = -18n
Then, add 8 to both sides (does not matter) = 0
Finally divide (answer is 0)

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3 years ago

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dimaraw [331]

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3 years ago

Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

D

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation $(x+0,y-2)$ will map vertices of the square into points

$(2,1)\rightarrow (2,-1);$
$(1,2)\rightarrow (1,0);$
$(2,3)\rightarrow (2,1);$
$(3,2)\rightarrow (3,0).$

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

$(2,-1)\rightarrow (2,3);$
$(1,0)\rightarrow (1,2);$
$(2,1)\rightarrow (2,1);$
$(3,0)\rightarrow (3,2)$

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation $(x+2,y-0)$ will map vertices of the square into points

$(2,1)\rightarrow (4,1);$
$(1,2)\rightarrow (3,2);$
$(2,3)\rightarrow (4,3);$
$(3,2)\rightarrow (5,2).$

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

$(4,1)\rightarrow (2,1);$
$(3,2)\rightarrow (3,2);$
$(4,3)\rightarrow (2,3);$
$(5,2)\rightarrow (1,2)$

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation $(x+3,y+3)$ will map vertices of the square into points

$(2,1)\rightarrow (5,4);$
$(1,2)\rightarrow (4,5);$
$(2,3)\rightarrow (5,6);$
$(3,2)\rightarrow (6,5).$

2nd transformation = reflection over y = -x + 7 will map vertices into points

$(5,4)\rightarrow (3,2);$
$(4,5)\rightarrow (2,3);$
$(5,6)\rightarrow (1,2);$
$(6,5)\rightarrow (2,1)$

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation $(x-3,y-3)$ will map vertices of the square into points

$(2,1)\rightarrow (-1,-2);$
$(1,2)\rightarrow (-2,-1);$
$(2,3)\rightarrow (-1,0);$
$(3,2)\rightarrow (0,-1).$

2nd transformation = reflection over y = -x + 2 will map vertices into points

$(-1,-2)\rightarrow (4,3);$
$(-2,-1)\rightarrow (3,4);$
$(-1,0)\rightarrow (2,3);$
$(0,-1)\rightarrow (3,2)$

These points are not the vertices of the initial square.

5 0

3 years ago

After traveling exactly half of its journey from station A to station B, a train was held up for 10 minutes. In order to arrive

Alex17521 [72]

Let's define variables:
s = original speed
s + 12 = faster speed
The time for the half of the route is:
60 / s
The time for the second half of the route is:
60 / (s + 12)
The equation for the time of the trip is:
60 / s + 60 / (s + 12) + 1/6 = 120 / s
Where,
1/6: held up for 10 minutes (in hours).
Rewriting the equation we have:
6s (60) + s (s + 12) = 60 * 6 (s + 12)
360s + s ^ 2 + 12s = 360s + 4320
s ^ 2 + 12s = 4320
s ^ 2 + 12s - 4320 = 0
We factor the equation:
(s + 72) (s-60) = 0
We take the positive root so that the problem makes physical sense.
s = 60 Km / h
Answer:
The original speed of the train before it was held up is:
s = 60 Km / h

8 0

3 years ago