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stellarik [79]
3 years ago
7

I'm stuck on graphing this quadratic equation:

%2B%204x%20%2B%203" id="TexFormula1" title="y \leqslant {x}^{2} + 4x + 3" alt="y \leqslant {x}^{2} + 4x + 3" align="absmiddle" class="latex-formula">

Mathematics
1 answer:
rewona [7]3 years ago
7 0
Your welcome I hope it is right

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Write the point-slope form of the equation of the line that passes through the points (-8, 2) and (1, -4). Include your work in
Tatiana [17]
Slope = (2+4)(-8 -1) = 6/-9 = - 2/3

<span>point-slope form of the equation
</span>
y - 2 = -2/3(x + 8)

3 0
3 years ago
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The slope f '(x) at each point (x, y) on a curve. y = f (x) is given along with a particular point (a, b) on the curve. Use this
Advocard [28]

Answer:

Step-by-step explanation:

Integrating each term with respect to x, we get:

            x^3          x^2

f(x) = 9--------- + 4------- - 4x + C

               3            2

We are told that if x = 0, f(x) = -7, and so C must equal - 7.  

The solution is

            x^3          x^2

f(x) = 9--------- + 4------- - 4x - 7,   or   f(x) = 3x^3 + 2x^2 - 4x - 7

               3            2

5 0
3 years ago
easy car rental charges $40 a day plus $.25 per mile race. rent a car charges $25 a day plus $.45 per mile. what number of miles
DiKsa [7]
A: cost= 3*55+.35m
B: cost=3*50+.40m

set the costs equal, and solve for m.
6 0
3 years ago
Find an equation of the tangent line to the hyperbola y = 3/x at the point (3, 1).
Vesna [10]
Let's find the derivative of that hyperbola in order to find a slope formula to help us with this equation.  We already have an x and y value.  The derivative is found this way:  y'= \frac{x(0)-3(1)}{x^2} so  y'=- \frac{3}{x^2}.  The derivative supplies us with the slope formula we need to write the equation.  Sub in the x value of 3 to find what the slope is: y'=- \frac{3}{3^2}=- \frac{3}{9}=- \frac{1}{3}.  So in our slope-intercept equation, x = 3, y = 1, and m = -1/3.  Use these values to solve for b.  1=- \frac{1}{3}(3)+b  so b = 2.  The equation, then, for the line tangent to that hyperbola at that given point is y=- \frac{1}{3}x+2
6 0
3 years ago
A grain silo has the shape of a right circular cylinder surmounted by a hemisphere. If the silo is to have a volume of 505π ft3,
Romashka-Z-Leto [24]

Answer:

radius   x  = 3 ft

height   h = 23,8  ft

Step-by-step explanation:

From problem statement

V(t)  = V(cylinder) + V(hemisphere)

let x be radius of base of cylinder (at the same time radius of the hemisphere)

and h the height of the cylinder, then:

V(c)  = π*x²*h       area of cylinder = area of base + lateral area

                                              A(c)  = π*x²  +   2*π*x*h

V(h) = (2/3)*π*x³   area of hemisphere   A(h)  =   (2/3)*π*x²

A(t)  =  π*x²  +   2*π*x*h +   (2/3)*π*x²

Now A as fuction of x    

total volume   505  = π*x²*h  +  (2/3)*π*x³

h = [505 - (2/3)* π*x³ ]  /2* π*x    

Now we have the expression for A as function of x

A(x) =  3π*x² + 2π*x*h     A(x) = 3π*x² + 505  - (2/3)π*x³

Taking derivatives both sides

A´(x) =  6πx -  2πx²              A´(x) =  0         6x  - 2x²  = 0

x₁  =  0  we dismiss

6 - 2x = 0

x = 3     and   h = [505 - (2/3)* π*x³]/2* π*x

h  =  (505 - 18.84) / 6.28*3

h  = 23,8  ft

7 0
3 years ago
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