Hello,

As 4²+(5/2)²=22.25=89/4 (Pytagorean theorem)
Perimeter of the triangle =2*√89 /2 +5 =5+√89
Lateral area: 18*(5+√89)
Total area= 90+18√89+ 4*5/2 *2 =110+18√89 ≈279.811..(ft²)

8 0

3 years ago

A line thru (5,1) with slope of 3.

VladimirAG [237]

<span>A line thru (5,1) with slope of 3

y - 1 = 3(x - 5)
y = 3x - 15 + 1
y = 3x - 14

hope it helps</span>

7 0

2 years ago

What is the value of a+ by when a= 7, b= and c= 1/2 ?

Marat540 [252]

Answer:

a = 7

b =?

c = 1/2

a + by = 0

7 + b × 1/2 = 0

7 + 1/2 b = 0

1/2 b = - 7

b = -7 × 2

b = - 14

a + by

= 7 + - 14 × 1/2

= 7 -14/2

= 7 - 7

= <u>0</u>

Step-by-step explanation:

Id.k if you forgot to write the value of b or not, since the value of b is not given, this will be ur answer.

3 0

3 years ago

Show that d^2y/dx^2=-2x/y^5, if x^3 + y^3=1

aalyn [17]

Answer:

y³ + x³ = 1

First, differentiate the first time, term by term:

${3y^{2}.\frac{dy}{dx} + 3x^{2}} = 0 \\\\{3y^{2}.\frac{dy}{dx} = -3x^{2}} \\\\\frac{dy}{dx} = \frac{-3x^{2}}{3y^{2}} \\\\\frac{dy}{dx} = \frac{-x^{2}}{y^{2}}$

↑ we'll substitute this later (4th step onwards)

Differentiate the second time:

$3y^{2}.\frac{dy}{dx} + 3x^{2} = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} + 6x = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{-x^{2} }{y^{2} })^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{x^{4} }{y^{4} }) = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + \frac{6x^{4} }{y^{3} } = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} = - 6x - \frac{6x^{4} }{y^{3} } \\\\$

$3y^{2}.\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{3y^{2}. y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 2xy^{3} - 2x^{4} }{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (y^{3} + x^{3})}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (1)}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x}{y^{5}}$

3 0

2 years ago