1 + sec^2(x)sin^2(x) = sec^2(x)
This becomes
1+tan^2(x) = sec^2(x) which is an identity
You could
1 + sin^2(x)/cos^2(x) = sec^2(x)
then
cos^2(x) + sin^2(x) = cos^2(x)sec^2(x)
1 = 1

4 0

3 years ago

Read 2 more answers

In care cadrane se afla graficul functiei ; f:(x)=121x

iragen [17]

Solución X = {0, 121}
espero que esto ayude

8 0

2 years ago

the relationship between the relative size of an earthquake, S, and the measure of the earthquake on the Richter scale, R, is gi

omeli [17]

Answer:

Relative Size, S, is 1584.89

Step-by-step explanation:

The equation is $R=log_{10}(S)$

Where,

R is the measurement in Richter Scale
S is the relative size of the earthquare

It is given that in Richter Scale, an earthquake measured 3.2, so $R=3.2$ .

They want to know relative size, S, so we put given information in equation and solve for S:

$R=log_{10}(S)\\3.2=log_{10}(S)$

Converting to exponential form, we have:

$3.2=log_{10}(S)\\S=10^{3.2}\\S=1584.8932$

Rounding to nearest hundredth, we have:

$S=1584.89$

8 0

2 years ago

Find the equation of a line that passes through the point (2,1) and has a gradient of 2. Leave your answer in the form ax + by =

Arte-miy333 [17]

Step-by-step explanation:

normally, if we have the gradient (or slope of inclination or change rate) and a point, we start using the point-slope form :

y - y1 = a(x - x1)

with (x1, y1) being a point on the line, and a being the slope of gradient (or ... however you want to call it).

y - 1 = 2(x - 2) = 2x - 4

y + 3 = 2x

2x - y = 3

8 0

1 year ago