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Svetlanka [38]
3 years ago
9

Apples are on sale at the grocery store at a price of 8 apples for $4.00. Maggie buys 12 apples. If she is charged the sale pric

e for apples, how much will Maggie pay?
Mathematics
1 answer:
Gnesinka [82]3 years ago
3 0

Answer:

$6

Step-by-step explanation:

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What is the simplest form of this expression
tekilochka [14]

Answer:

In general, an expression is in simplest form when it is easiest to use. Example, this: 5x + x − 3. Is simpler as: 6x − 3. Common ways to help you simplify: • Combine Like Terms.Step-by-step explanation:sorry if this is isin't right

7 0
3 years ago
2.The number of grams A of a certain radioactive substance present at time, in yearsfrom the present, t is given by the formulaA
professor190 [17]

Answer:

Given that,

The number of grams A of a certain radioactive substance present at time, in years

from the present, t is given by the formula

A=45e^{-0.0045(t)}

a) To find the initial amount of this substance

At t=0, we get

A=45e^{-0.0045(0)}A=45e^0

We know that e^0=1 ( anything to the power zero is 1)

we get,

A=45

The initial amount of the substance is 45 grams

b)To find thehalf-life of this substance

To find t when the substance becames half the amount.

A=45/2

Substitute this we get,

\frac{45}{2}=45e^{-0.0045(t)}

\frac{1}{2}=e^{-0.0045(t)}

Taking natural logarithm on both sides we get,

\ln (\frac{1}{2})=-0.0045(t)^{}(-1)\ln (\frac{1}{2})=0.0045(t)\ln (\frac{1}{2})^{-1}=0.0045(t)\ln (2)=0.0045(t)0.6931=0.0045(t)t=\frac{0.6931}{0.0045}t=154.02

Half-life of this substance is 154.02

c) To find the amount of substance will be present around in 2500 years

Put t=2500

we get,

A=45e^{-0.0045(2500)}A=45e^{-11.25}A=45\times0.000013=0.000585A=0.000585

The amount of substance will be present around in 2500 years is 0.000585 grams

4 0
1 year ago
Determine two pairs of polar coordinates for the point (5, 5) with 0° ≤ θ < 360°.
inna [77]

Answer:

First part

The answer is (5 square root 2, 45°), (-5 square root 2, 225°) ⇒ answer (d)

Second part

The equation in standard form for the hyperbola is y²/81 - x²/19 = 1 ⇒ answer(b)

Step-by-step explanation:

First part:

* Lets study the Polar form and the Cartesian form

- The important difference between Cartesian coordinates and

  polar coordinates:

# In Cartesian coordinates there is exactly one set of coordinates

  for any given point.

# In polar coordinates there is an infinite number of coordinates

   for a given point. For instance, the following four points are all

   coordinates for the same point.

# In the polar the coordinates the origin is called the pole, and

  the x axis is called the polar axis.

# The angle measurement θ can be expressed in radians

   or degrees.

- To convert from Cartesian Coordinates (x , y) to

  Polar Coordinates (r , θ)

# r = ± √(x² + y²)

# θ = tan^-1 (y / x)

* Lets solve the problem

- The point in the Cartesian coordinates is (5 , 5)

∵ x = 5 and y = 5

∴ r = ± √(5² + 5²) = ± √50 = ± 5√2

∴ tanФ = (5/5) = 1

∵ tanФ is positive

∴ Angle Ф could be in the first or third quadrant

∵ Ф = tan^-1 (1) = 45°

∴ Ф in the first quadrant is 45°

∴ Ф in the third quadrant is 180 + 45 = 225°

* The answer is (5√2 , 45°) , (-5√2 , 225°)

Second part:

* Lets study the standard form of the hyperbola equation

- The standard form of the equation of a hyperbola with  

  center (0 , 0) and transverse axis parallel to the y-axis is

  y²/a² - x²/b² = 1, where

• the length of the transverse axis is 2a

• the coordinates of the vertices are (0 , ±a)

• the length of the conjugate axis is 2b

• the coordinates of the co-vertices are (±b , 0)

•      the coordinates of the foci are (0 , ± c),  

• the distance between the foci is 2c, where c² = a² + b²

* Lets solve our problem

∵ The vertices are (0 , 9) and (0 , -9)

∴ a = ± 9 ⇒ a² = 81

∵ The foci at (0 , 10) , (0 , -10)

∴ c = ± 10

∵ c² = a² + b²

∴ (10)² = (9)² + b² ⇒ 100 = 81 + b² ⇒ subtract 81 from both sides

∴ b² = 19

∵ The equation is  y²/a² - x²/b² = 1

∴  y²/81 - x²/19 = 1

* The equation in standard form for the hyperbola is y²/81 - x²/19 = 1

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