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Marysya12 [62]
3 years ago
11

The following graph is of an exponential function of the form y=a*bx.

Mathematics
1 answer:
FromTheMoon [43]3 years ago
5 0

Answer:

WAIT no one has ever answered this?? what....

Step-by-step explanation:

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2x-2y=-6

-2y= -2x-6

y=x+3

Y-intercept= 3

X-intercept= -3

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The first three terms of a geometric sequence are as follows. -256, -64, -16. Find the next two terms of this sequence. Give exa
umka2103 [35]
Hi,


The next two are:

- 16 \div 4 =  - 4 \\  - 4 \div 4 =  - 1

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Hope this helps.
r3t40
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3 years ago
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

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2 years ago
A salesperson earned a salary of $12000 per year plus 5% commission on sales of $100000. What is his total income
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If im doing right....

100,000*.5=50,000

Then 12,000+50,000=62,000 per year.

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