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Leno4ka [110]
1 year ago
15

Please help, functions

Mathematics
1 answer:
Mila [183]1 year ago
4 0

Step-by-step explanation:

equate the equation to zero

x³-6x+11x-6=0

if we replace x with 1

it will result to zero

1³-6+11-6=0

using long Division

as shown in the picture

obtain

x²-5x+6

using factorisation

x²-5x+6

(x²-2x)-(3x+6)

x(x-2)-3(x-2)

x=2

x=3

x=1

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What is 5 1/2 +2 1/7 due in one hour help please
nika2105 [10]

Answer:

7 9/14

Step-by-step explanation:

1/2=7/14

1/7=2/14

7/14+2/14=9/14

5+2=7

6 0
2 years ago
Read 2 more answers
Find the value of the expression 0.5^10/0.5^7
SIZIF [17.4K]

Answer:

The solution is:

  • \frac{0.5^{10}}{0.5^7}=0.125

Step-by-step explanation:

Given the expression

0.5^{10}\div 0.5^7

Solving

\frac{0.5^{10}}{0.5^7}

\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}

so

\frac{0.5^{10}}{0.5^7}=0.5^{10-7}

\mathrm{Subtract\:the\:numbers:}\:10-7=3

=0.5^3

=0.125        ∵ 0.5^3=0.125

Therefore, the solution is:

\frac{0.5^{10}}{0.5^7}=0.125

7 0
3 years ago
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Which expression is 8y - 5x + 13y equivalent to after being simplified?
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3 years ago
równanie zmiany prędkość autokaru poruszającego się po prostym odcinku szosy i rozpoczynającego hamowanie od szybkości 20m/s ma
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Answer:

s = 22.5 m

Step-by-step explanation:

the equation for the speed change of a coach moving along a straight section of the road and starting braking at a speed of 20 m / s has the form v (t) = 25-5t. Using integral calculus, determine the coach's braking distance.

v (t) = 25 - 5 t

at t = 0 , v = 20 m/s

Let the distance is s.

s =\int v(t) dt\\\\s =\int (25 - 5t)dt\\\\s= 25 t - 2.5 t^2 \\

Let at t = t, the v = 20

So,

20 = 25 - 5 t

t = 1 s

So, s = 25 x 1 - 2.5 x 1 = 22.5 m

3 0
2 years ago
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