All categories

1 year ago

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

You might be interested in

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

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

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

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

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

$=0.5^3$

$=0.125$ ∵ $0.5^3=0.125$

Therefore, the solution is:

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

7 0

3 years ago

Can someone please help with this question? Please A cylindrical tank has a radius of 15 ft. and a height of 45 ft. How many cub

Reil [10]

I did the problem myself and my teacher always said thatif you don't get the exact answer then round to the closest one. The answer that i got was 31,808.6, the closest one to that would be letter C.

8 0

3 years ago

Which expression is 8y - 5x + 13y equivalent to after being simplified?

Thepotemich [5.8K]

You combine like terms like 8y+13y which gives you 21y and don't forget the -5x so your simplified equation would end up being A. 21 - 5x

7 0

3 years ago

równanie zmiany prędkość autokaru poruszającego się po prostym odcinku szosy i rozpoczynającego hamowanie od szybkości 20m/s ma

Rasek [7]

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