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3 years ago

Can you help with Issue A?

Mathematics

1 answer:

defon3 years ago

7 0

Answer:

$16,532.98

Step-by-step explanation:

Her last deposit = 500, second last = 500 x 1.05, third last = 500 x 1.05^2................ first deposit = 500 x 1.05^19

=500 +500 x 1.05 + 500 x 1.05^2 + 500 x 1.05^19

= 500(1.05^20-1) divided by (1.05-1)

= 16532.98

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Help I will give brainliest

scoray [572]

Answer:

43/70

Step-by-step explanation:

Total number of times spun = 14 + 12 + 16 + 15 + 13 = 70

Total number of times spun that yielded a odd number (1,3,5) = 14 + 16 + 13 = 43

The probability of spinning an odd number = total number of spins that yielded 1,3,5 ÷ total number of spins yielding all values from 1 to 5

= 43 / 70

8 0

3 years ago

ANSWER ASAP PLEASE AND SHOW YOUR WORK!!!!

IgorLugansk [536]

Answer:

f(-x)= -f(x) so, the given function $f(x)= x^3-2x$ is Odd

Step-by-step explanation:

We need to determine if the function $f(x)= x^3-2x$ is Even, Odd, or Neither.

Determining if a function f(x) is even or not we put x=-x and check

If f(-x)=f(x) the function is even

if f(-x)= -f(x) the function is odd

So, Putting x =-x and determining if the given function $f(x)= x^3-2x$ is Even, Odd, or Neither.

$f(x)= x^3-2x\\Put \ x=-x\\f(-x)=(-x)^3-2(-x)\\f(-x)=-x^3+2x\\f(-x)=-(x^3-2x)\\f(-x)=-f(x)$

As f(-x)= -f(x) so, the given function $f(x)= x^3-2x$ is Odd

5 0

3 years ago

What is the greatest common factor of 20x^4 and 6x^3 X^3 X^4 2x^3 2x^4

Lesechka [4]

The GCF of 20x^4 and 6x^3 would be " 2x^3 "

3 0

3 years ago

Read 2 more answers

I want yo nkow how to simplify

xxMikexx [17]

Remove parentheses by multiplying factors.
use exponent rules to remove parentheses in terms with exponents.
combine like terms by adding coefficients.
combine the constants.

5 0

3 years ago

What is 5.316 - 1.942 (show ur work)

Fiesta28 [93]

Answer:

3.374

Step-by-step explanation:

$\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}$

$\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}$

$\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}$

$\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4$

$\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}$

$\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\mathrm{The\:bottom\:number\:is\:larger\:than\:the\:upper\:number.\:\:Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}$

$\mathrm{The\:top\:digit\:is\:not\:bigger\:than\:the\:bottom\:one.\:\:Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}$

$\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\mathrm{Borrow\:}1\mathrm{\:from\:}5\mathrm{.\:\:The\:remainder\:is\:}4$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&10&\:\:&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&3&1&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{\:\:}&\:\:&\:\:&\:\:&4\end{matrix}}$

$\mathrm{Add\:}1\mathrm{\:ten\:to\:}3:\quad \:10+3=13$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\mathrm{Borrow\:}1\mathrm{\:from\:}13\mathrm{.\:\:The\:remainder\:is\:}12$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\mathrm{Add\:}1\mathrm{\:ten\:to\:}1:\quad \:10+1=11$

$\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:11-4=7$

$\mathrm{Place\:the\:decimal\:point\:in\:the\:answer\:directly\:below\:the\:decimal\:points\:in\:the\:terms}$

$\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:4-1=3$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}$

$=3.374$

Hence the correct answer is 3.374

7 0

2 years ago