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Mars2501 [29]
2 years ago
10

3x² - 2x + 1 = [?] x=4

Mathematics
2 answers:
astraxan [27]2 years ago
7 0

Answer:

3(4)² - 2(4) + 1 = 48 - 8 +1 = 41

antiseptic1488 [7]2 years ago
5 0

x = 4

{3x}^{2}  - 2x + 1 \\  \\ 3 \times  {4}^{2}  - 2 \times 4 + 1 \\  \\ 3 \times 16 -8  + 1 \\  \\ 48 - 8 + 1 \\  \\ 41.

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Ranger says the expression is not in its simplest form. He thinks the two terms need to be combined. Is Ranger right that 8x − 2
ladessa [460]

Answer:

4(2x - 5)

Step-by-step explanation:

Rewrite "20" as 4 * 5

Rewrite "8" as 4 * 2

=4 * 2x - 4 * 5

Factor out common term "4"

=4(2x - 5) <------ Your answer.

5 0
3 years ago
Read 2 more answers
Factor completely: <br> x^3 - x^2 - 6x
PSYCHO15rus [73]

Answer:

x(x-3)(x+2)

Step-by-step explanation:

1. Find the GCF (Greatest Common Factor)

GCF=x

2. Factor out the GCF and simplify.

x(x^{2} -x-6)

3. Factor x^{2} -x-6.

Which two numbers add up to -1 and multiply to -6?

<u>-3 and 2</u>

Rewrite the expression using the numbers above.

(x-3)(x+2)

4. Done!

x(x-3)(x+2)

4 0
3 years ago
£8.40 is shared equally between three brothers. How much does each brother receive?
alexandr1967 [171]
8.40/3 = £2.80 each
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7 0
3 years ago
Calculate pt3 such that a line from pt1 to pt3 is perpendicular to the line from pt1 to pt2, and the distance between pt1 and pt
Leni [432]
Let the point_1 = p₁ = (1,4)
and      point_2 = p₂ = (-2,1)
and      Point_3 = p₃ = (x,y)

The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)

The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d = \sqrt{Δx^2+Δy^2}
d₁ = \sqrt{(-2-1)^2+(1-4)^2}
d₂ = \sqrt{(x-1)^2+(y-4)^2}
d₁ = d₂
∴ \sqrt{(-2-1)^2+(1-4)^2} = \sqrt{(x-1)^2+(y-4)^2} ⇒⇒ eliminating the root
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
 (x-1)²+(y-4)² = 18
from equatoin (1)  y-4 = 1-x
∴(x-1)²+(1-x)² = 18            ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 = \pm \sqrt{9} = \pm 3
∴ x = 4 or x = -2
∴ y = 1 or y = 7

Point_3 = (4,1)  or  (-2,7)












8 0
4 years ago
Find an equation for the line that passes through the point (-3, 6), parallel to the line through the points (0,-7) and (4,-15).
Vitek1552 [10]
Broken down into steps:

1.  Find the slope of the line segment that connecs the points (0,-7) and (4,-15).

2.  Start with the point-slope formula for the equation of a straight line:

y-b = m(x-a), where the given point is (a,b).  Borrow the value of m that you calculated in (1), above, and insert it into this point-slope formula.
Finish up by subst. the x- and y-values in (-3,6) into this formula.

Done!

You could, of course, solve this result for y if you wished.
8 0
4 years ago
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