Interest = Pin = principal * interest rate * number of years

Let interest rate be i, then
600*i*2+150*i*4=90
1200i+600i = 90
i=90/(1200+600)=90/1800=1/20=0.05 or 5%

8 0

3 years ago

NEED HELP ASAP !! WILL MARK BRAINLEAST<br> SHOW WORK TOO !!!

irinina [24]

Step-by-step explanation:

$(1 + \cos(x) )(1 - \cos(x) )$

$1 - \cos {}^{2} (x)$

$= \sin {}^{2} (x)$

$\frac{1}{ \cot {}^{2} (x) } - \frac{1}{ \cos {}^{2} (x) }$

$\frac{1}{ \frac{ \cos {}^{2} (x) }{ \sin { }^{2} (x) } } - \frac{1}{ \cos {}^{2} (x) }$

$\frac{1}{ \cot {}^{2} (x) } - \frac{ \csc {}^{2} (x) {} }{ \cot {}^{2} (x) }$

$\frac{ - \cot {}^{2} (x) }{ \cot {}^{2} (x) }$

$- 1$

$\sec {}^{2} ( \frac{\pi}{2} - x ) )( \sin {}^{2} (x) - \sin {}^{4} (x))$

$( \csc {}^{2} (x) )( \sin {}^{2} (x) - \sin {}^{4} (x )$

$\csc {}^{2} (x) ( \frac{1}{ \csc {}^{2} (x) } - \frac{1}{ \csc {}^{4} (x) } )$

$1 - \frac{1}{ \csc {}^{2} (x) }$

$1 - \sin {}^{2} (x)$

$\cos {}^{2} (x)$

3 0

2 years ago

Solve log9^5 =7- 3X by graphing. Which equation should be graphed?

olya-2409 [2.1K]

Answer:

y^1 = log5/log9 & y^2 = 7 - 3x

3 0

3 years ago

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