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3 years ago

Why do whole numbers raised to an exponent get greater well fractions raised to an exponent get smaller

1 answer:

4 0

Because when you raise an exponent on a fraction, it is basically dividing into itself. Imagine 1/3 multiplied my 3 (Ans is 1), the fractions aren't whole numbers so when they multiply into another number, it uses division

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A ball is thrown with an initial speed of 17 mph in a direction that makes an angle of 22 degrees with the positive x-axis. Expr

Agata [3.3K]

Answer:

The speed of the ball in terms of i and j is (15.76 i, 6.37 j)

Step-by-step explanation:

Given;

velocity of the ball, v = 17 mph

angle of inclination, θ = 22⁰

The horizontal component of the velocity is calculated as;

$V_x = Vcos\theta\\\\V_x = 17cos(22)\\\\V_x = 15.76 \ i$

The vertical component of the velocity is calculated as;

$V_y = Vsin\theta\\\\V_y = 17sin(22)\\\\V_y = 6.37 \ j$

Therefore, the speed of the ball in terms of i and j is (15.76 i, 6.37 j)

7 0

2 years ago

The measure of an angle is nine times the measure of its complement. find the measure of each angle.

andrew11 [14]

Two angles are complementary when they add up to 90 degrees

smaller angle = x
larger angle = 9x

x + 9x = 90
10x = 90
x = 90/10
x = 9° ← smaller angle

larger angle = 9x = 9 * 9 = 81°

5 0

3 years ago

Which number comes next in this series: 2, 3, 4, 8, 9, 10, 20, 21, 22,?

Mekhanik [1.2K]

44

The pattern is +1,+1,x2

3 0

3 years ago

Linda fills the rectangular prism above (volume=56 cubed inches) with cubes that have side lengths of 1/3 in. How many cubes doe

Pani-rosa [81]

The answer to your question is 2 2/3

5 0

3 years ago

Harmonic mean if a number h such that (h-a)/(b-h)=a/b Prove h is H(a,b) iff satisfies either relation a. (1/a)-(1/h) = (1/h) - (

Fudgin [204]

A.

$\displaystyle\frac1a-\frac1h=\frac1h-\frac1b$
$\implies\displaystyle\frac{h-a}{ah}=\frac{b-h}{bh}$
$\implies\displaystyle\frac{h-a}{b-h}=\frac{ah}{bh}$
$\implies\displaystyle\frac{h-a}{b-h}=\frac ab$

b.

$\displaystyle h=\frac{2ab}{a+b}$
$\displaystyle\implies\frac{h-a}{b-h}=\frac{\frac{2ab}{a+b}-a}{b-\frac{2ab}{a+b}}$
$\displaystyle\implies\frac{h-a}{b-h}=\frac{2ab-a(a+b)}{b(a+b)-2ab}$
$\displaystyle\implies\frac{h-a}{b-h}=\frac{ab-a^2}{b^2-ab}$
$\displaystyle\implies\frac{h-a}{b-h}=\frac{a(b-a)}{b(b-a)}$
$\displaystyle\implies\frac{h-a}{b-h}=\frac ab$

The other direction can be proved by following the manipulations in the reverse order.

7 0

3 years ago