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3 years ago

What is 4 to the 2/3 power

1 answer:

8 0

$\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
-------------------------------\\\\
4^{\frac{2}{3}}\implies \sqrt[3]{4^2}\implies \sqrt[3]{(2^2)^2}\implies \sqrt[3]{2^4}\implies \sqrt[3]{2^{3+1}}\implies \sqrt[3]{2^3\cdot 2}
\\\\\\
2\sqrt[3]{2}$

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What is the first term of the fibonacci like sequence whose second term is 4 and whose fith term is 22?

bagirrra123 [75]

Answer: -2

Step-by-step explanation:

The information in the question can be used to form an equation which goes thus:

2nd term = a + d = 4 ...... i

5th term = a + 4d = 22 ....... ii

From equation i

a = 4 - d ........ iii

Put equation iii into ii

a + 4d = 22

(4 - d) + 4d = 22

4 - d + 4d = 22

4 + 3d = 22

3d = 22 - 4

3d = 18

d = 18/3

d = 6

Commons difference is 6

Since a + d = 4

a + 6 = 4

a = 4 - 6

a = -2

The first term is -2

3 0

3 years ago

Can somebody help me please.

serg [7]

Answer:

0,-8

-6,0

Step-by-step explanation:

5 0

2 years ago

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Olce Question What type of notation uses these brackets {}? A. Interval B. Set

Sergio [31]

Answer:

Interval?

Step-by-step explanation:

5 0

3 years ago

An object launched straight up at a speed of 29.4 meters per second has a height, h, in meters of h , t seconds after the object

Nostrana [21]

Answer:

h = 44.06 meters (maximum height)

the time the object takes to complete this whole path is 6 seconds, this is why the time at which the object reaches its maximum height will between 0 and 6 seconds

Step-by-step explanation:

To solve this question, we need to first recognize that this is a constant acceleration problem, specifically, it can be thought of as a projectile motion problem.

Recall, the equations of motion:

$1) v^2 - v_0^2 = 2a(s - s_0)\\2) s = v_0^2 + \frac{1}{2} at^2\\3) v = v_0 + at$

What do we already know?

$v_0 = 29.4 ms^-1$
The launch is straight up
$a = -9.81 ms^-2$ this is the gravitational acceleration $g$

$s_0 = 0 m$ , since our reference point is at s = 0, (the ground)

We can use use the Eq(1):

we know that when any object is launched up, at maximum height its velocity is going to be zero, $v = 0 ms^-2$

$v^2 - v_0^2 = 2a(s - s_0)\\0^2 - (29.4)^2 = 2(-9.81)(s- 0)\\s = 44.06 m$

this is the maximum height!

Why does t have to between zero and six?

We can answer this using a bit visualization, if you think about the second equation

$s = v_0 t - \frac{1}{2}at^2\\ s = 29.4t - 4.905t^2$

this is the equation of the whole trajectory that object makes.

and if you solve this by making $s = 0$ , you will get the times at which the object was at the ground. the times will be 0s and 5.99s.

so the amount of time the object takes to go through this whole path is 6 seconds and this why the object will only reach its maximum height in between this time interval.

hope this helps :)

5 0

3 years ago

Kristen invests $ 5745 in a bank. The bank 6.5% interest compounded monthly. How long must she leave the money in the bank for i

podryga [215]

Answer:

1. 10.7 years

2. 17.0 years

3. 2nd option

Step-by-step explanation:

Use formula for compounded interest

$A=P\cdot \left(1+\dfrac{r}{n}\right)^{nt},$

where

A is final value, P is initial value, r is interest rate (as decimal), n is number of periods and t is number of years.

In your case,

1. P=$5745, A=2P=$11490, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then

$11490=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},\\ \\2=(1.0054)^{12t},\\ \\12t=\log_{1.0054}2,\\ \\t=\dfrac{1}{12}\log_ {1.0054}2\approx 10.7\ years.$

2. P=$5745, A=3P=$17235, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then

$17235=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},$

$3=(1.0054)^{12t},$

$12t=\log_{1.0054}3,$

$t=\dfrac{1}{12}\log_{1.0054}3\approx 17.0\ years.$

3. 1 choice: P=$5745, n=12 (compounded monthly), r=0.065 (6.5%), t=5 years and A is unknown. Then

$A=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12\cdot 5},\\ \\A=5745\cdot (1.0054)^{60}\approx \$7936.39.$

2 choice: P=$5745, n=4 (compounded monthly), r=0.0675 (6.75%), t=5 years and A is unknown. Then

$A=5745\cdot \left(1+\dfrac{0.0675}{4}\right)^{4\cdot 5},\\ \\A=5745\cdot (1.0169)^{20}\approx \$8032.58.$

The best will be 2nd option, because $8032.58>$7936.39

5 0

3 years ago

Read 2 more answers