What's -8 + (-1 1/8)

1 answer:

6 0

I think the answer is 9.375 if it needs to be in decimal form. If it needs to be in fraction form it would be improper, 9 3/8 would be the fraction. Hope it helps!

You might be interested in

Please please help!!

shepuryov [24]

Answer:

$\large\boxed{x=0\ and\ x=\pi}$

Step-by-step explanation:

$\tan^2x\sec^2x+2\sec^2x-\tan^2x=2\\\\\text{Use}\ \tan x=\dfrac{\sin x}{\cos x},\ \sec x=\dfrac{1}{\cos x}:\\\\\left(\dfrac{\sin x}{\cos x}\right)^2\left(\dfrac{1}{\cos x}\right)^2+2\left(\dfrac{1}{\cos x}\right)^2-\left(\dfrac{\sin x}{\cos x}\right)^2=2\\\\\left(\dfrac{\sin^2x}{\cos^2x}\right)\left(\dfrac{1}{\cos^2x}\right)+\dfrac{2}{\cos^2x}-\dfrac{\sin^2x}{\cos^2x}=2$

$\dfrac{\sin^2x}{(\cos^2x)^2}+\dfrac{2-\sin^2x}{\cos^2x}=2\\\\\text{Use}\ \sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{2-(1-\cos^2x)}{\cos^2x}=2\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{2-1+\cos^2x}{\cos^2x}=2\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{1+\cos^2x}{\cos^2x}=2$

$\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{(1+\cos^2x)(\cos^2x)}{(\cos^2x)^2}=2\qquad\text{Use the distributive property}\\\\\dfrac{1-\cos^2x+\cos^2x+\cos^4x}{\cos^4x}=2\\\\\dfrac{1+\cos^4x}{\cos^4x}=2\qquad\text{multiply both sides by}\ \cos^4x\neq0\\\\1+\cos^4x=2\cos^4x\qquad\text{subtract}\ \cos^4x\ \text{from both sides}\\\\1=\cos^4x\iff \cos x=\pm\sqrt1\to\cos x=\pm1\\\\ x=k\pi\ for\ k\in\mathbb{Z}\\\\\text{On the interval}\ 0\leq x$

4 0

2 years ago

8 (Picture) CONVERGENT AND DIVERGENT SERIES PLEASE HELP!!

djverab [1.8K]

Answer:

The correct answer choice is: B

Step-by-step explanation:

Culture A

Initial number of cells = 1000

The number of cells is raised by 200 each day and every day thereafter.

So, the number of cells raised in one month(30 days) $=200\times 30=6000$

Thus, the total number of cells in culture A at the end of one month will be: $1000+6000=7000$

Culture B

Initial number of cells = 1000

The number of cells is increased by 8% each day and every day thereafter.

Formula for exponential growth is: $A=P(1+r)^t$ , where $P=$ Initial amount, $A=$ Final amount, $r=$ Rate of growth in decimal form and $t=$ Time duration.