An equation for water level in june for pensacola as a function of time (t) is f(t) = 5 cos pi/6 t + 7.
Which equation of cos show period amplitude ?
The equation given below show aplitude and period
where A = amplitude,
b = 2 pi/Period,
Period = 12 hrs,
c = midline,
x = t and y = f(t)
We have to find the amplitude
<h3>What is the formula for the amplitude?</h3>
Therefore, the an equation for water level in june for pensacola as a function of time (t)
To learn more about the function of time visit:
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Answer:
x = 12
Step-by-step explanation:
Recall: the secant-tangent rule states that when a secant and a tangent meet at an external point of a circle, the product of the secant and the external segment is equal to the square of the tangent segment
(x)(3) = 6² (secant-tangent rule)
3x = 36
Divide both sides by 3
3x/3 = 36/3
x = 12
To answer this, you will use the area of 900 square yards to determine the distances between the bases. Each side of the square is 30 yards, so it will be 30 yards from 1st to home and from 1st to 2nd.
The distance from home to 2nd is a diagonal in the square (the hypotenuse).
You will use the Pythagorean Theorem to find this distance.
a^2 + b^2 = c^2
30^2 + 30 ^2 = c^2
900 + 900 + c^2
1800 = c^2
The square root of 1800 is approximately 42.4 yards.
The ball travels approximately 42.4 yards.
1. Start with ΔCIJ.
- ∠HIC and ∠CIJ are supplementary, then m∠CIJ=180°-7x;
- the sum of the measures of all interior angles in ΔCIJ is 180°, then m∠CJI=180°-m∠JCI-m∠CIJ=180°-25°-(180°-7x)=7x-25°;
- ∠CJI and ∠KJA are congruent as vertical angles, then m∠KJA =m∠CJI=7x-25°.
2. Lines HM and DG are parallel, then ∠KJA and ∠JAB are consecutive interior angles, then m∠KJA+m∠JAB=180°. So
m∠JAB=180°-m∠KJA=180°-(7x-25°)=205°-7x.
3. Consider ΔCKL.
- ∠LFG and ∠CLM are corresponding angles, then m∠LFG=m∠CLM=8x;
- ∠CLM and ∠CLK are supplementary, then m∠CLM+m∠CLK=180°, m∠CLK=180°-8x;
- the sum of the measures of all interior angles in ΔCLK is 180°, then m∠CKL=180°-m∠CLK-m∠LCK=180°-(180°-8x)-42°=8x-42°;
- ∠CKL and ∠JKB are congruent as vertical angles, then m∠JKB =m∠CKL=8x-42°.
4. Lines HM and DG are parallel, then ∠JKB and ∠KBA are consecutive interior angles, then m∠JKB+m∠KBA=180°. So
m∠KBA=180°-m∠JKB=180°-(8x-42°)=222°-8x.
5. ΔABC is isosceles, then angles adjacent to the base are congruent:
m∠KBA=m∠JAB → 222°-8x=205°-7x,
7x-8x=205°-222°,
-x=-17°,
x=17°.
Then m∠CAB=m∠CBA=205°-7x=86°.
Answer: 86°.
