Answer:
The answer w'll be obtained using formulas
cos(a+b) = cosacosb - sinasinb
cos(a-b) = cosacosb + sinasinb
Step-by-step explanation:
Using the trigonometric formula of addition and subtraction of cosine
cos(a+b) = cosacosb - sinasinb
cos(a-b) = cosacosb + sinasinb
w'll get the desired answer.
To be solve
L.H.S = R.H.S
sinasinb = (cos(a-b)-cos(a+b)/2
as we know that <u><em>cos(a+b) = cosacosb - sinasinb</em></u>
sinasinb = (cos(a-b) - (cosacosb -sinasinb))/2
as we know that <u><em>cos(a-b) = cosacosb + sinasinb</em></u>
sinasinb = ((cosacosb + sinasinb) - (cosacosb -sinasinb))/2
sinasinb = (cosacosb + sinasinb - cosacosb + sinasinb)/2
sinasinb = (2sinasinb)/2
sinasinb = sinasinb
hence L.H.S = R.H.S
.50 x 2 < .20 x 100
1 < 20
50% of 2 is not greater than 20% of 100.
The answer:
the main rules of the use of logarithm are
loga[a] = 1
loga[AxB] =loga[A] +loga[B] for all value positive of A and B
loga[A/B] = loga[A] - loga[B] for all value positive of A and B
in our case, <span>log8 4a (b-4/c4)
so it is equivalent to </span>log8 4a + <span>log8(b-4/c4)
and since </span>loga[A/B] = loga[A] l - oga[B] , log8(b-4/c4) =log8(b-4) - log8(c4)
the possible expression:
log8 4a (b-4/c4) = log8 4a + log8(b-4) - log8(c4)
The temperature that accurately describes the time that would have spent cooling is 300.
<h3>How to solve for the cooling time.</h3>
We have the function as f(t) = 349.2(0.98)t
We have to solve for all of the values in the option
When t = 0
This would give us an infinite value for t hence it is not the answer.
When t = 100
we have 61.6
When t = 300
we have f(t) = 7.5
WHen t = 400
f(t) = -6.7
From the solved values the accurate one that has to talk about the time spent cooling is 300 at 7.5
<h3>Complete question</h3>
The function f(t) = 349.2(0.98)t models the relationship between t, the time an oven spends cooling and the temperature of the oven. For which temperature will the model most accurately predict the time spent cooling? 0 100 300 400
Read more on the temperature here:
brainly.com/question/12564447
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