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3 years ago

Help me please its super easy

Mathematics

2 answers:

yanalaym [24]3 years ago

6 0

Your answer would be the first option 10^5.

Hope this helped.

7nadin3 [17]3 years ago

6 0

10^5

That is because since 10 is being times to itself 5 times, then the exponent would be 10^5

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Solve for A if is 3.14 and R = 3

gavmur [86]

Answer:

A = 28.26

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Step-by-step explanation:

Step 1: Define

Equation: A = πr²

Given: π = 3.14, r = 3

Step 2: Evaluate

Substitute: A = 3.14(3)²
Exponents: A = 3.14(9)
Multiply: A = 28.26

5 0

3 years ago

An airliner maintaining a constant elevation of 2 miles passes over an airport at noon traveling 500 mi/hr due west. At 1:00 PM,

butalik [34]

Answer:

$\frac{ds}{dt}\approx 743.303\,\frac{mi}{h}$

Step-by-step explanation:

Let suppose that airliners travel at constant speed. The equations for travelled distance of each airplane with respect to origin are respectively:

First airplane

$r_{A} = 500\,\frac{mi}{h}\cdot t\\r_{B} = 550\,\frac{mi}{h}\cdot t$

Where t is the time measured in hours.

Since north and west are perpendicular to each other, the staight distance between airliners can modelled by means of the Pythagorean Theorem:

$s=\sqrt{r_{A}^{2}+r_{B}^{2}}$

Rate of change of such distance can be found by the deriving the expression in terms of time:

$\frac{ds}{dt}=\frac{r_{A}\cdot \frac{dr_{A}}{dt}+r_{B}\cdot \frac{dr_{B}}{dt}}{\sqrt{r_{A}^{2}+r_{B}^{2}} }$

Where $\frac{dr_{A}}{dt} = 500\,\frac{mi}{h}$ and $\frac{dr_{B}}{dt} = 550\,\frac{mi}{h}$ , respectively. Distances of each airliner at 2:30 PM are:

$r_{A}= (500\,\frac{mi}{h})\cdot (1.5\,h)\\r_{A} = 750\,mi$

$r_{B}=(550\,\frac{mi}{h} )\cdot (1.5\,h)\\r_{B} = 825\,mi$

The rate of change is:

$\frac{ds}{dt}=\frac{(750\,mi)\cdot (500\,\frac{mi}{h} )+(825\,mi)\cdot(550\,\frac{mi}{h})}{\sqrt{(750\,mi)^{2}+(825\,mi)^{2}} }$

$\frac{ds}{dt}\approx 743.303\,\frac{mi}{h}$

6 0

3 years ago

Answers for both boxes please

adelina 88 [10]

Answer:

3,0

Step-by-step explanation:

7 0

3 years ago

A store sold 50 copies of a magazine for $150. Each copy of the magazine costs the same. Which equation and set of ordered pairs

Goryan [66]

Answer:

Option a. Y=3x

Step-by-step explanation:

Let us use cross multiplication method.

Let the cost of 1 magazine be x.

No. of copies Cost

1)50 $150

2)1 x

50x=150 x 1 equation(1)

x=150/50

x=$3

Now see equation (1),

150=50x

150=50 x 3

Here let us represent the cost as y and no. of copies as x.

Y=3x

Therefore, a. Y=3x is the right answer.

Thank you!

8 0

3 years ago

50 point available!!!!!!

g100num [7]

Answer:

2 miles is your answer hope this helps!!!!!!

Step-by-step explanation:

7 0

3 years ago

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