The parent function is $f(x) = 2^x$ . Plugging in x = 0 leads to y = 1. So the point (0,1) is on the f(x) curve. Going from (0,1) to (0,-2) is a vertical shift of 3 units downward. To represent this shift, we tack on a "-3" at the end of the f(x) function.

$g(x) = f(x) - 3\\\\g(x) = 2^x - 3$

You could look at other points as well, but I find working with x = 0 is easiest.

As a check, plugging x = 0 into g(x) leads to...

$g(x) = 2^x - 3\\\\g(0) = 2^0 - 3\\\\g(0) = 1 - 3\\\\g(0) = -2$

This confirms our answer.

4 0

2 years ago

Help I am very bad at math I need answers by tomorrow if you can help please do

d1i1m1o1n [39]

Answer:

Step-by-step explanation:

(a - b)(a +b) = a² - b²

1 - Sin² A = Cos² A

$LHS = \frac{1}{1- Sin A} + \frac{1}{1 + Sin A}\\\\= \frac{1*(1 + Sin A)}{(1- Sin A)(1 + Sin A)} + \frac{1*(1- Sin A)}{(1 + Sin A)(1- Sin A)}\\\\= \frac{1 + Sin A+ 1 - Sin A}{1^{2}- Sin^{2} A}\\\\= \frac{2}{1 - Sin^{2} A}\\\\= \frac{2}{Cos^{2} A}\\\\= 2 Sec^{2} A$

2) Sec² A - Tan² A = 1

$LHS = \frac{1}{Sec A - Tan A}\\\\=\frac{1*(Sec A + Tan A)}{(Sec A - Tan A)(Sec A + Tan A)}\\\\=\frac{Sec A + Tan A}{Sec^{2} A - Tan^{2} A}\\\\=\frac{Sec A + Tan A }{1}\\\\= Sec A + Tan A = RHS\\\\\\$

3) LHS = Cosec² A + Cot² A

= Cosec² A + Cosec² A - 1

= 2Cosec² A - 1 = RHS

$4) LHS = \frac{Sec A}{Cos A}- \frac{Tan A}{Cot A}\\\\ = Sec A*\frac{1}{Cos A}-Tan A*\frac{1}{Cot A}\\\\ = Sec A * Sec A - Tan A * Tan A\\\\= Sec^{2} A - Tan^{2} A \\\\= 1$

3 0

2 years ago

What is 16+x=12 and -3x=15

horrorfan [7]

-4 and -5
because to get 16 to 12 you have to subtract 4 but it wants you to add so you put a negative sign infront of it.

and a negative multiply by a negative equals a positive so -3×-5=15

5 0

3 years ago

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