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aleksandrvk [35]
3 years ago
15

Factor the following expression, and verify that the factored expression is equivalent to the original: 16x²-8x-3

Mathematics
1 answer:
vlabodo [156]3 years ago
5 0

Answer:

Solved  (x-3/4)(x+1/4)

Step-by-step explanation:

16x²-8x-3=0

we can find the factors of the above quadratic equation using shridharacharya  formula which given as

x=\frac{-b\pm \sqrt{b^2-4ac} }{2a}

a= 16, b= -8 and c= -3 putting these values in above equation we can find the factors of x as

x=\frac{8\pm \sqrt{8^2-4\times16\times(-3)} }{2\times16}

x= 3/4 and -1/4

now  16x²-8x-3 can also be written as (x-3/4)(x+1/4) as 3/4 and 1/4 are the roots of the equation.

on solving expression is equivalent to the original: 16x²-8x-3 we will again obtain  16x²-8x-3. Hence it is varified that the factored expression is equivalent to the original: 16x²-8x-3

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anygoal [31]

Answer:

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5 0
3 years ago
Given f(x)=-2x^3 + 3x^2 , find the equation of that tangent line of f at the point where x=2.
trapecia [35]

Step 1: Find f'(x):

f'(x) = -6x^2 + 6x

Step 2: Evaluate f'(2) to find the slope of the tangent line at x=2:

f'(2) = -6(2)^2 + 6(2) = -24 + 12 = -12

Step 3: Find f(2), so you have a point on y=f(x):

f(2) = -2·(2)^3 + 3·(2)^2 = -16 + 12 = -4

So, you have the point (2,-4) and the slope of -12.

Step 4: Find the equation of your tangent line:

Using point-slope form you'd have: y + 4 = -12 (x - 2)

That is the equation of the tangent line.

If your teacher is picky and wants slope-intercept, solve that for y to get:

y = -12 x + 20

5 0
2 years ago
Cmon this is my last question help me
Lelechka [254]
Blue Box: 6.03
Orange Box: -3
5 0
3 years ago
Read 2 more answers
Linear or nonlinear tables?
Oxana [17]

Table 1 and Table 2 are linear tables while table 3 is a non-linear table.

Step-by-step explanation:

Step 1; If y's value is dependent on the value of x, it is a linear relationship i.e. as the value of x varies, y also varies correspondingly. So x is independent while y is dependent on x i.e. the value of y depends on the value of x. We need to determine if there is a linear relationship between the values of x and y in the given tables.

Step 2; In table 1, the values of y are squares of the values of x i.e. 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25. So the relationship is y = x². So table 1 is a linear table i.e. values which exist in a linear relationship.

Step 3; In table 2, the values of y are obtained by summing two times the value of x and 4.

when x = 1, y = 2(1) + 4 = 6,

when x = 2, y = 2(2) + 4 = 8,

when x = 3, y = 2(3) + 4 = 10,

when x = 4, y = 2(4) + 4 = 12,

when x = 5, y = 2(5) + 4 = 14.

So y = 2x + 4, so table 2 is also a linear table.

Step 4; There is no relationship between the values of y and x. For the negative values of x, y = -x while for positive values of x, y = x, there is no constant relationship between the values of y and x. So table 3 is a non-linear table.

4 0
3 years ago
Evaluate x/y + 3z^2 for x= 2/3, y= 6/7, and z=3
prohojiy [21]

x/y + 3z^2 for x= 2/3, y= 6/7, and z=3

=2/3  /  6/7  + 3(3^2)

=2/3 * 7/6 + 3(9)

= 7/9 + 27

= 7/9 + 243/9

= 250/9

answer is a. 250/9

6 0
3 years ago
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