Round off 11.25 to the nearest hundred to the nearest cent

Mathematics

1 answer:

TEA [102]3 years ago

5 0

11.25 is rounded to 11.30

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I need help bruv. For the function y=x^2 - 5x - 6 .

lara [203]

Answer:

a. Vertex = (2.5,-12.25)

b. y-intercept = (0,-6)

c. x-intercept = (6,0) ; (-1,0)

Step-by-step explanation:

Your equation is written as:

$y = ax^{2} + bx +c$

The easiest way to find the vertex is writing the equation this way:

$y = a(x-h)^{2}+k$

Being the vertex (h,k)

So first complete the square

$y = x^{2} -5x -6 \\y = x^{2} - 2(2.5x) -6 \\y = x^{2} - 2(2.5x) +2.5^{2} - 2.5^{2} -6 \\y = (x^{2} -2(2.5x) + 2.5^{2}) -6.25 - 6 \\y = (x-2.5)^{2} -12.25$

Vertex : (2.5,-12.25)

To find the y-intercept you need to replace the equation when x = 0 and get y

$y = (0)^{2} - 5(0) -6\\y = - 6$

To find the x-intercept you need to replace the equation when y = 0 and get x

$0 = x^{2} - 5x -6 \\0 = (x-6)(x+1)\\x = 6\\x = -1$

4 0

3 years ago

Simplify (9k6 + 8k4 - 6k2)(4k2 - 5)

Tasya [4]

Answer:

$36k^8-13k^6-64k^4+30k^2$

Step-by-step explanation:

$36k^8-13k^6-64k^4+30k^2\\9k^6*(4k^2-5)+8k^4*(4k^2-5)-6k^2*(4x^2-5)\\36k^8-45k^6+8k^4*(4k^2-5)-6k^2*(4x^2-5)\\36k^8-45k^6+32k^6-40k^4-6k^2*(4x^2-5)\\36k^8-45k^6+32k^6-40k^4-24k^4+30k^2\\36k^8-13k^6-40k^4-24k^4+30k^2\\36k^8-13k^6-64k^4+30k^2$