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3 years ago

The perimeter of a square is the product of four and the length of a side. Which of the following funct

1 answer:

5 0

<h3><u>The function used to find the perimeter of a square, represented by y, given the length of a side, represented by x is: y = 4x</u></h3>

<em><u>Solution:</u></em>

Let "y" be the perimeter of square

Let the length of side be "x"

From given,

The perimeter of a square is the product of four and the length of a side

Which means,

$\text{Perimeter of square } = 4 \times \text{ length of side }$

Substituting the given variables,

$y = 4 \times x\\\\y = 4x$

Thus perimeter of square is y = 4x

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I'm stuck on this one:( help me please

ZanzabumX [31]

Answer:

Step-by-step explanation:

f-1(x) is the inverse of f(x)

to find the inverse let’s set the equation up like this

y = 4x -2

now we’ll switch x and y

x = 4y -2

and solve for y

x + 2 = 4y

x/4 + 2/4 = y

the inverse is

y = x/4 + 1/2

this equation is written in y = mx + b form

where b = y intercept

so the y intercept is 1/2

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3 years ago

Tristan had 95 inches of ribbon.He cut off 27.50 inches. How many inches of ribbon remain

tatiyna

Answer:

67.50 inches

Step-by-step explanation:

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3 years ago

Given the two points what is the slope of the line?<br> (4,2) (10.4)

zlopas [31]

$m = \frac{y2-y1}{x2-x1}\\ m=\frac{4-2}{10-4} \\m = \frac{2}{6} \\\\m=\frac{1}{3}$

Therefore, the slope of the line is 1/3. In case if you want the equation too.

$y=mx+b\\y=\frac{1}{3}x+b\\ 2=\frac{1}{3}(4)+b\\ 2=\frac{4}{3}+b\\ b=-\frac{4}{3}+2\\ b=-\frac{4}{3} +\frac{6}{3}\\ b=\frac{2}{3}$

Therefore, the equation is y=1/3x+2/3

6 0

3 years ago

SOMEONE PLEASE HELP ME WITH THIS AND EXPLAIN EACH ONE AND HOW YOU DID IT PLEASE MY TEACHER SUCK AND ILL GIVE YOU BRAINLY IF U GE

frutty [35]

Answer:

pretty sure its a

Step-by-step explanation:

I do not have a explanation sorry bro

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3 years ago

Let f(x)=x^3-x-1

alexira [117]

$f(x) = x^3 - x - 1$

To find the gradient of the tangent, we must first differentiate the function.

$f'(x) = \frac{d}{dx}(x^3 - x - 1) = 3x^2 - 1$

The gradient at x = 0 is given by evaluating f'(0).

$f'(0) = 3(0)^2 - 1 = -1$

The derivative of the function at this point is negative, which tells us <em>the function is decreasing at that point</em>.

The tangent to the line is a straight line, so we will have a linear equation of the form y = mx + c. We know the gradient, m, is equal to -1, so

$y = -x + c$

Now we need to substitute a point on the tangent into this equation to find c. We know a point when x = 0 lies on here. To find the y-coordinate of this point we need to evaluate f(0).

$f(0) = (0)^3 - (0) - 1 = -1$

So the point (0, -1) lies on the tangent. Substituting into the tangent equation:

$y = -x + c \\\\ -1 = -(0) + c \\\\ -1 = c \\\\ \text{Equation of tangent is } \boxed{y = -x - 1}$

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3 years ago