All categories

2 years ago

I NEED HELP PLEASE, IM STUCK ON MOST OF THEM

Mathematics

1 answer:

azamat2 years ago

6 0

Answer: look it up

Step-by-step explanation: Lollll

You might be interested in

ULLIVLY/1/2/3/assessment

zvonat [6]

<h3>The worth after 4 years is $ 680.24</h3>

Solution:

The formula for compound interest, including principal sum, is:

$A = p(1+\frac{r}{n})^{nt}$

Where,

A = the future value of the investment

P = the principal investment amount

r = the annual interest rate (decimal)

n = the number of times that interest is compounded per unit t

t = the time the money is invested

From given,

n = 1 ( since interest is compounded annually)

p = 500

t = 4

$r = 8 \% = \frac{8}{100} = 0.08$

Substituting the values we get,

$A = 500(1+ \frac{0.08}{1})^{1 \times 4}\\\\A = 500(1.08)^4\\\\A = 500 \times 1.36048896\\\\A = 680.24448 \approx 680.24$

Thus the worth after 4 years is $ 680.24

3 0

3 years ago

Which equation finds the length of side c of this right triangle

NeX [460]

Answer:

Find The Length Of The Side Of A Right Triangle : Example Question #1. Explanation: The Pythagorean Theorem gives us a2 + b2 = c2 for a right triangle, where c is the hypotenuse and a and b are the smaller sides.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

(8 x 3) + (8 x 2) = 8 x (blank + blank)

kakasveta [241]

Step-by-step explanation:

(8 x 3) + (8 x 2) = 8 x (3 + 2)

24 + 16 = 8 * 5

40 = 40

Hope it will help :)❤

8 0

2 years ago

VEEL

Andre45 [30]

Answer:

$a_n=-3(3)^{n-1}$ ; {-3,-9, -27,- 81, -243, ...}

$a_n=-3(-3)^{n-1}$ ; {-3, 9,-27, 81, -243, ...}

$a_n=3(\frac{1}{2})^{n-1}$ ; {3, 1.5, 0.75, 0.375, 0.1875, ...}

$a_n=243(\frac{1}{3})^{n-1}$ ; {243, 81, 27, 9, 3, ...}

Step-by-step explanation:

The first explicit equation is

$a_n=-3(3)^{n-1}$

At n=1,

$a_1=-3(3)^{1-1}=-3$

At n=2,

$a_2=-3(3)^{2-1}=-9$

At n=3,

$a_3=-3(3)^{3-1}=-27$

Therefore, the geometric sequence is {-3,-9, -27,- 81, -243, ...}.

The second explicit equation is

$a_n=-3(-3)^{n-1}$

At n=1,

$a_1=-3(-3)^{1-1}=-3$

At n=2,

$a_2=-3(-3)^{2-1}=9$

At n=3,

$a_3=-3(-3)^{3-1}=-27$

Therefore, the geometric sequence is {-3, 9,-27, 81, -243, ...}.

The third explicit equation is

$a_n=3(\frac{1}{2})^{n-1}$

At n=1,

$a_1=3(\frac{1}{2})^{1-1}=3$

At n=2,

$a_2=3(\frac{1}{2})^{2-1}=1.5$

At n=3,

$a_3=3(\frac{1}{2})^{3-1}=0.75$

Therefore, the geometric sequence is {3, 1.5, 0.75, 0.375, 0.1875, ...}.

The fourth explicit equation is

$a_n=243(\frac{1}{3})^{n-1}$

At n=1,

$a_1=243(\frac{1}{3})^{1-1}=243$

At n=2,

$a_2=243(\frac{1}{3})^{2-1}=81$

At n=3,

$a_3=243(\frac{1}{3})^{3-1}=27$

Therefore, the geometric sequence is {243, 81, 27, 9, 3, ...}.

6 0

3 years ago

What does k equal if 485= 1/((k)(5.99x10^-3))

Verizon [17]

$k\ne 0$

$485=\frac{1}{(k)(5.99*10^{-3})}\\\\ \frac{485}{1}= \frac{1}{0.00599k}\\\\ 485*0.00599k=1\\\\ k=\frac{1}{485*0.00599}\\\\k\approx 0.3442$

7 0

2 years ago