!)) can you guess what these three digts are

A swimming pool is 10 m long, 8 m wide and 1.5 m deep. The inside of the pool

alisha [4.7K]

80 1.5 because 10 times 8 equals 80 plus 1.5 equals 80.1.5

6 0

3 years ago

Read 2 more answers

A pyramid is 1 3 inch taller than it is wide. Let h=height. Let w=width. Complete the table, and graph h=w+ 1 3.

ivann1987 [24]

Answer:

Step-by-step explanation:

o,9.o,0

5 0

3 years ago

Please Help fast!!!!! I'll give 30 pts.!

Zinaida [17]

The best buy is from dealership B because dealership B has a lesser base price

<h3>The add-on options</h3>

For the purpose of this exercise, the add-on options to select are:

Sunroof
Keyless entry

<h3>Dealership A</h3>

The total amount of the add-on options is:

Total = $500 + $1500

Total = $2000

<u>The inequality</u>

This is represented using:

Base price * (1 + Tax) + Fees + Add-on total ≤ 20000

So, we have:

(1 + 6.75%) * x + 751.56 + 2000 ≤ 20000

<u>Solve for x</u>

We have:

(1 + 6.75%) * x + 751.56 + 2000 ≤ 20000

Evaluate the like terms

(1 + 6.75%) * x ≤ 17248.44

Divide both sides by (1 + 6.75%)

x ≤ 16157.79

The solution means that the base price of the car cannot exceed $16157.78 for dealership A

<u>The graph of the possible base price</u>

See attachment

<h3>Dealership B</h3>

The total amount of the add-on options is:

Total = $750 + $1000

Total = $1750

<u>The inequality</u>

This is represented using:

Base price * (1 + Tax) + Fees + Add-on total ≤ 20000

So, we have:

(1 + 6.75%) * x + 1524.72 + 1750 ≤ 20000

<u>Solve for x</u>

We have:

(1 + 6.75%) * x + 1524.72 + 1750 ≤ 20000

Evaluate the like terms

(1 + 6.75%) * x ≤ 16725.28

Divide both sides by (1 + 6.75%)

x ≤ 15667.71

The solution means that the base price of the car cannot exceed $15667.71 for dealership B

<u>The graph of the possible base price</u>

See attachment

<h3>Where to buy from</h3>

Buy from dealership B because dealership B has a lesser base price

Read more about inequality at:

brainly.com/question/25275758

#SPJ1

3 0

2 years ago

Prove that 1/sec0-tan0= sec0+tan0

Leviafan [203]

Answer:

Step-by-step explanation:

$\text{LHS}=\frac{1}{\sec \theta-\tan \theta}\\=\frac{\sec \theta+\tan \theta}{(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)}\\=\frac{\sec \theta+\tan \theta}{\sec^{2} \theta-\tan^{2} \theta}\\=\frac{\sec \theta+\tan \theta}{1}\\=\sec \theta+\tan \theta\\=\text{RHS}$

7 0

2 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

3 years ago