3 years ago

Find f'(x). ,f(x)=radical3x+1

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

5 0

Answer:

Step-by-step explanation:

hello :

here is an solution :

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Simplify each of the following expressions to find the length of the tightrope

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What is the length ?

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Complete the work to find the dimensions of the rectangle. x(x – 3) = 10 x2 – 3x = 10 x2 – 3x – 10 = 10 – 10 (x + 2)(x – 5) = 0

Norma-Jean [14]

Answer:

$Length =5$

$Width = 2$

Step-by-step explanation:

Given

$Length =x$

$Width = x -3$

$Area = 10$

$x(x-3) =10$

See comment for missing part of the question

Required

Complete the expression to determine the dimension of a rectangle

We have:

$x(x-3) =10$

Open bracket

$x^2 -3x = 10$

Equate to 0

$x^2 -3x - 10 =0$

Expand

$x^2 + 2x - 5x - 10 = 0$

Factorize

$x(x + 2) - 5(x + 2) = 0$

Factor out x + 2

$(x - 5)(x + 2) = 0$

Solve for x

$x - 5 = 0$ or $x + 2 = 0$

$x = 5$ or $x = -2$

The value of x cannot be negative

So:

$x = 5$

Recall that:

$Length = x$

$Width = x - 3$

So:

$Length =5$

$Width = 2$ ---- i.e. 5 - 3

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Which relation is not a function?

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The Answer is A.

Repeating number is not a function. It's ok to have repeating number in Y or the range but not the X which is the Domain.

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What is the midpoint of the vertical line segment graphed below? (2, 4), (2, -9) A. (4, -5). B. (2, -5/2). C. (2, -5). D. (4, 5/

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Answer:

B. $(2, -\frac{5}{2})$

Step-by-step explanation:

Given:

(2, 4) and (2, -9)

Required:

Midpoint of the vertical line with the above endpoints

Solution:

Apply the midpoint formula, which is:

$M(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$

Where,

(2, 4) = (x_1, y_1)

(2, -9) = (x_2, y_2)

Plug in the values into the equation:

$M(\frac{2 + 2}{2}, \frac{4 + (-9)}{2})$

$M(\frac{4}{2}, \frac{-5}{2})$

$M(2, -\frac{5}{2})$

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Find f'(x). ,f(x)=radical3x+1​

Find f'(x). ,f(x)=radical3x+1