All categories

3 years ago

(5x-2)^2 help asap pleaseee

Mathematics

1 answer:

omeli [17]3 years ago

5 0

Answer:

25x² − 20x + 4

Explanation:

(5x − 2)²

= (5x + −2)(5x + −2)

= (5x)(5x) + (5x)(−2) + (−2)(5x) + (−2)(−2)

= 25x² − 10x − 10x + 4

= 25x² − 20x + 4

You might be interested in

How to do expanded form and word for for 119 000 003

kiruha [24]

Expanded- 100000000+10000000+9000000+4

5 0

3 years ago

Please answer! <br> Solve for X

Sedbober [7]

Answer:

x = 3

Step-by-step explanation:

We can solve this by using ratios of similar triangles

3 x

------ = ------------

3+9 12

3 x

------ = ------------

12 12

x must equal 3 to keep the ratios the same

4 0

3 years ago

Read 2 more answers

3xy - 15x + y ^ 2 - c In the polynomial above, c is a constant. If the polynomial is divisible by y - 5 , what is the value of c

Natalija [7]

Answer:

Step-by-step explanation:

Synthetic division shows the remainder to be 25-c. In order for that remainder to be zero, the value of c must be 25.

4 0

3 years ago

The perimeter of a rectangle is 46 m if the with were doubled and on length were increased by 5 their perimeter would be 66 what

Kamila [148]

Answer:

18 and 5

Step-by-step explanation:

if 'x' - is the length and 'y' - the width of the rectangle, then

2(x+y)=46 - perimeter of the original rectangle and 2(2x+(y+5))=66 - the perimeter of the new rectangle.

$\left \{ {{x+y=23} \atop {2x+y=28}} \right. \=> \ \left \{ {{x=5} \atop {y=18}} \right.$

Finally, the length is x=5 and the width is y=18.

8 0

3 years ago

Complete parts a through c for the given function.

victus00 [196]

Answer:

a. The critcal points are at

$x=0,-5,3$

b. Then, $x = -5$ is a maximum and $x=3$ is a minimum

c. The absolute minimum is at $x = 3$ and the absolute maximum is at $x = -5.$

Step-by-step explanation:

(a)

Remember that you need to find the points where

$f'(x)=0$

Therefore you have to solve this equation.

$20x^4 + 40x^3 - 300x^2 = 0$

From that equation you can factor out $20x^2$ and you would get

$20x^2 ( x^2 +2x - 15) = 0$

And from that you would have $20x^2 = 0$ , so $x = 0$ .

And you would also have $x^2 +2x-15 = 0$ .

You can factor that equation as $x^2 +2x -15 = (x+5)(x-3) = 0$

Therefore $x=-5 , x=3$ .

So the critcal points are at

$x=0,-5,3$

Remember that a function has a maximum at a critical point if the second derivative at that point is negative. Since

$f''(x) = 80x^3 + 120x^2 -600x\\f''(-5) = 80(-5)^3 + 120(-5)^2 -600(-5) = -4000 < 0\\\\f''(3) = 80(3)^3 + 120(3)^2 -600(3) = 1440 > 0 \\$

Then, $x = -5$ is a maximum and $x=3$ is a minimum

The absolute minimum is at $x = 3$ and the absolute maximum is at $x = -5.$

6 0

3 years ago

Read 2 more answers