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4 years ago

What is the solution of -5m+10(6=8m)

Mathematics

1 answer:

hodyreva [135]4 years ago

4 0

Answer:

m= 2 or 0.75

Step-by-step explanation:

$- 5m + 10(6 = 8m) \\ - 30m + 60 = - 40{m}^{2} + 80m \\ 60 = - 40{m}^{2} + 80m + 30m \\$

$60 = - 40{m}^{2} + 110m \\0 = - 40{m}^{2$ } + 110m - 60 \\ 0 = - (40 {m}^{2} - 110m + 60) \\ 0 = - (5m - 10)(8m - 6) [/tex]

split into two equations:

$0 = - (5m - 10) \\ 0 = - 5m + 10 \\ - 10 = - 5m \\ m = 2$

and

$0 = 8m - 6 \\ 6 = 8m \\ m = 0.75$

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How many real-number solutions does the equation have?

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Answer:

Option b. Two solutions

Step-by-step explanation:

In order to find how many real number solutions the equation has we have to solve it

Given equation: -4x² + 10x + 6 = 0

taking 2 common from the equation

2(-2x² + 5x + 3) = 0

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taking minus sign common from the above equation

2x² - 5x - 3 = 0

We will solve this equation by factorization in such a way that the sum of two factors is equal to -5x and the product is -6x²

2x² - 6x + x - 3 = 0

taking common above

2x(x-3) + 1(x-3) = 0

taking (x-3) common

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3 years ago

What would y=x^2 +x+ 2 be in vertex form

balu736 [363]

Answer:

$y = (x + \frac{1}{2} )^{2} + \frac{7}{4}$

Step-by-step explanation:

$y = {x}^{2} + x + 2$

We can covert the standard form into the vertex form by either using the formula, completing the square or with calculus.

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

The following equation above is the vertex form of Quadratic Function.

Vertex — Formula

$h = - \frac{b}{2a} \\ k = \frac{4ac - {b}^{2} }{4a}$

We substitute the value of these terms from the standard form.

$y = a {x}^{2} + bx + c$

$h = - \frac{1}{2(1)} \\ h = - \frac{ 1}{2}$

Our h is - 1/2

$k = \frac{4(1)(2) - ( {1})^{2} }{4(1)} \\ k = \frac{8 - 1}{4} \\ k = \frac{7}{4}$

Our k is 7/4.

Vertex — Calculus

We can use differential or derivative to find the vertex as well.

$f(x) = a {x}^{n}$

Therefore our derivative of f(x) —

$f'(x) = n \times a {x}^{n - 1}$

From the standard form of the given equation.

$y = {x}^{2} + x + 2$

Differentiate the following equation. We can use the dy/dx symbol instead of f'(x) or y'

$f'(x) = (2 \times 1 {x}^{2 - 1} ) + (1 \times {x}^{1 - 1} ) + 0$

Any constants that are differentiated will automatically become 0.

$f'(x) = 2 {x}+ 1$

Then we substitute f'(x) = 0

$0 =2x + 1 \\ 2x + 1 = 0 \\ 2x = - 1 \\x = - \frac{1}{2}$

Because x = h. Therefore, h = - 1/2

Then substitute x = -1/2 in the function (not differentiated function)

$y = {x}^{2} + x + 2$

$y = ( - \frac{1}{2} )^{2} + ( - \frac{1}{2} ) + 2 \\ y = \frac{1}{4} - \frac{1}{2} + 2 \\ y = \frac{1}{4} - \frac{2}{4} + \frac{8}{4} \\ y = \frac{7}{4}$

Because y = k. Our k is 7/4.

From the vertex form, our vertex is at (h,k)

Therefore, substitute h = -1/2 and k = 7/4 in the equation.

$y = a {(x - h)}^{2} + k \\ y = (x - ( - \frac{1}{2} ))^{2} + \frac{7}{4} \\ y = (x + \frac{1}{2} )^{2} + \frac{7}{4}$

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