Does anyone mind helping me with this?

Mathematics

1 answer:

ZanzabumX [31]4 years ago

3 0

4.5% = 0.045
so
$39.98 x 0.045 = $1.80
answer
sale tax = $1.80

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Solve the following equation for "p": -13p = -377

saul85 [17]

Divide each term by -13 and simplify.
P= 29
Hope this helped!

4 0

3 years ago

Which is an equation in Point-Slope form for the given point and slope

svetoff [14.1K]

Answer:

The equation of this line is y - 7 = 4(x + 3)

Step-by-step explanation:

To find this, start with the base form of point-slope form.

y - y1 = m(x - x1)

Now put the slope in for m and the two coordinates in for (x1, y1).

y - 7 = 4(x + 3)

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3 years ago

Give the slope-intercept form of the equation of the line that is perpendicular to 7x + 3y = 18 and contains P(6, 8).

Olegator [25]

Rearrange the equation into slope intercept form

7x+3y=18

3y=-7x+18

y=-7x/3+18/3 so the slope of the reference line is -7/3

To be perpendicular to this line the slope must be the negative reciprocal of this...mathematically m1*m2=-1. So the slope of our perpendicular line is:

-7m/3=-1

-7m=-3

m=3/7 so our line so far is:

y=3x/7+b, using point (6,8) we can solve for b, the y-intercept

8=3(6)/7+b

8=18/7+b

56/7-18/7=b

38/7=b so our line is:

y=3x/7+38/7 or if you prefer

y=(3x+38)/7

4 0

3 years ago

Use f’( x ) = lim With h ---> 0 [f( x + h ) - f ( x )]/h to find the derivative at x for the given function. 5-x²

beks73 [17]

<h2>Answer:</h2>

The derivative of the function f(x) is:

$f'(x)=-2x$

<h2>Step-by-step explanation:</h2>

We are given a function f(x) as:

$f(x)=5-x^2$

We have:

$f(x+h)=5-(x+h)^2\\\\i.e.\\\\f(x+h)=5-(x^2+h^2+2xh)$

( Since,

$(a+b)^2=a^2+b^2+2ab$ )

Hence, we get:

$f(x+h)=5-x^2-h^2-2xh$

Also, by using the definition of f'(x) i.e.

$f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

Hence, on putting the value in the formula:

$f'(x)= \lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-(5-x^2)}{h}\\\\\\f'(x)=\lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-5+x^2}{h}\\\\i.e.\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2-2xh}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2}{h}+\dfrac{-2xh}{h}\\\\f'(x)=\lim_{h \to 0} -h-2x\\\\i.e.\ on\ putting\ the\ limit\ we\ obtain:\\\\f'(x)=-2x$