All categories

2 years ago

Ryan invested $5,900 in an account paying an interest rate of 5% compounded

Mathematics

2 answers:

Phoenix [80]2 years ago

7 0

Answer:

10214.51

Step-by-step explanation:

vladimir1956 [14]2 years ago

4 0

Answer:

10214.51

Step-by-step explanation:

5900 * (1+ (0.05/12))^(12*11)

please give brainliest

You might be interested in

Divide. Round your answer to the nearest tenth.

nika2105 [10]

The answer to your question is. 103.2

3 0

2 years ago

3/4<br> What fraction is equivalent to the fraction above.

SSSSS [86.1K]

Answer: 6/8, 12/16, 15/20, 18/24, are all equal

Step-by-step explanation:

4 0

3 years ago

Evaluate the expression when x= -3.<br><br> x2-5x-8

never [62]

Step-by-step explanation:

x²-5x-8

When x=-3, then

(-3)²-5×(-3)-8
9+15-8
24-8
16

4 0

3 years ago

Read 2 more answers

Rewrite 9cos 4x in terms of cos x.

rosijanka [135]

$\bf \qquad \textit{Quad identities}\\\\
sin(4\theta )=
\begin{cases}
8sin(\theta )cos^3(\theta )-4sin(\theta )cos(\theta )\\
4sin(\theta )cos(\theta )-8sin^3(\theta )cos(\theta )
\end{cases}
\\\\\\
cos(4\theta)=8cos^4(\theta )-8cos^2(\theta )+1\\\\
-------------------------------\\\\
9cos(4x)\implies 9[8cos^4(x)-8cos^2(x)+1]
\\\\\\
72cos^4(x)-72cos^2(x)+9$

---------------------------------------------------------------------------

as far as the previous one on the 2tan(3x)

$\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\qquad tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}\\\\
-------------------------------\\\\$

$\bf 2tan(3x)\implies 2tan(2x+x)\implies 2\left[ \cfrac{tan(2x)+tan(x)}{1-tan(2x)tan(x)}\right]
\\\\\\
2\left[ \cfrac{\frac{2tan(x)}{1-tan^2(x)}+tan(x)}{1-\frac{2tan(x)}{1-tan^2(x)}tan(x)}\right]\implies 2\left[ \cfrac{\frac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}}{\frac{1-tan(x)-2tan^3(x)}{1-tan^2(x)}} \right]
\\\\\\$

$\bf 2\left[ \cfrac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}\cdot \cfrac{1-tan^2(x)}{1-tan(x)-2tan^3(x)} \right]
\\\\\\
2\left[ \cfrac{3tan(x)-tan^3(x)}{1-tan^2(x)-2tan^3(x)} \right]\implies \cfrac{6tan(x)-2tan^3(x)}{1-tan^2(x)-2tan^3(x)}$

4 0

3 years ago

Someone help please

dalvyx [7]

Answer:

Step-by-step explanation:

the 2 angles shown are complementary meaning that they will add up to equal 90

x+71=90

subtract 71 from each side x=19

5 0

3 years ago

Read 2 more answers