All categories

2 years ago

Helpppp]ppppppppppppp

Mathematics

2 answers:

Natali5045456 [20]2 years ago

7 0

Answer:

Step-by-step explanation:

madam [21]2 years ago

3 0

Answer:

I think it's b or d

Step-by-step explanation:

You might be interested in

What is the length of the altitude of the equilateral triangle below?

Mice21 [21]

B. 3 √3

The shortcuts for a 30 60 90 triangle is that the:

Hypotenuse is 2 times the short leg

The long leg is √3 times the short leg.

Making it 3 √3

Another way is by the Pythagorean Theorem.

a^2+b^2=c^2

3^2+b^2=6^2

9+b^2=36

Subtract 9 from both sides

b^2=27

Square root both sides

b=5.2 or the √27

√27 can be simplified more

An equation that equals to 27 that has a perfect square is 9*3

√9* √3

The perfect square of 9 equals to 3

So 3 √3

4 0

2 years ago

Which value is in the domain of f(x)?

ahrayia [7]

Answer:

Step-by-step explanation:

$f(x)=-2x+3, 0$

4 0

3 years ago

Read 2 more answers

the graph of the equation below is a circle. what is the length of the radius of the cirlce? (x+5)2+(y+7)2=21^2

zloy xaker [14]

the radii would equal 7.5

8 0

2 years ago

By 8:00 a.m., the snack bar had made $3,256 in sales. By noon, the snack bar’s total sales were $8,821. On average, how much mon

dimulka [17.4K]

8821 - 3256 = 5565....amount made between 8 a.m. and noon
8 a.m. to 12: a.m. = 4 hrs

5565 / 4 = 1391.25 <=== what they took in each hr between 8 and noon

3 0

2 years ago

Read 2 more answers

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