The length of his path to the nearest tenth of a foot is 37.7 ft option fourth is correct.
<h3>What is the Pythagoras theorem?</h3>
The square of the hypotenuse in a right-angled triangle is equal to the sum of the squares of the other two sides.
We have:
Frankie wants to build a path from one corner of his yard to the opposite corner. his yard measures 20 ft. x 32 ft.
From the Pythagoras theorem:
Perpendicular(P) = 32 ft
Base(B) = 20 ft
H = √(32²+20²)
H = √1424
H = 37.7 ft
Thus, the length of his path to the nearest tenth of a foot is 37.7 ft option fourth is correct.
Learn more about Pythagoras' theorem here:
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Answer:
Step-by-step explanation:
if we have 5^22 and 5^26
5^26= 5^22 x5^4
5^26 is bigger by the number 5^4
Answer:
common ratio: 1.155
rate of growth: 15.5 %
Step-by-step explanation:
The model for exponential growth of population P looks like: 
where 
is the population at time "t",
is the initial (starting) population
is the common ratio,
and 
is the rate of growth
Therefore, in our case we can replace specific values in this expression (including population after 12 years, and initial population), and solve for the unknown common ratio and its related rate of growth:
![P(t)=P_i(1+r)^t\\13000=2300*(1+r)^{12}\\\frac{13000}{2300} = (1+r)^12\\\frac{130}{23} = (1+r)^{12}\\1+r=\sqrt[12]{\frac{130}{23} } =1.155273\\](https://tex.z-dn.net/?f=P%28t%29%3DP_i%281%2Br%29%5Et%5C%5C13000%3D2300%2A%281%2Br%29%5E%7B12%7D%5C%5C%5Cfrac%7B13000%7D%7B2300%7D%20%3D%20%281%2Br%29%5E12%5C%5C%5Cfrac%7B130%7D%7B23%7D%20%3D%20%281%2Br%29%5E%7B12%7D%5C%5C1%2Br%3D%5Csqrt%5B12%5D%7B%5Cfrac%7B130%7D%7B23%7D%20%7D%20%3D1.155273%5C%5C)
This (1+r) is the common ratio, that we are asked to round to the nearest thousandth, so we use: 1.155
We are also asked to find the rate of increase (r), and to express it in percent form. Therefore we use the last equation shown above to solve for "r" and express tin percent form:
So, this number in percent form (and rounded to the nearest tenth as requested) is: 15.5 %
Answer:
45
Step-by-step explanation:
Two tangents drawn to a circle from an outside point form arcs and an angle, and this formula shows the relation between the angle and the two arcs.
m<EYL = (1/2)(m(arc)EVL - m(arc)EHL) Eq. 1
The sum of the angle measures of the two arcs is the angle measure of the entire circle, 360 deg.
m(arc)EVL + m(arc)EHL = 360
m(arc)EVL = 360 - m(arc)EHL Eq. 2
We are given this:
m<EYL = (1/3)m(arc)EHL Eq. 3
Substitute equations 2 and 3 into equation 1.
(1/3)m(arc)EHL = (1/2)[(360 - m(arc)EHL) - m(arc)EHL]
Now we have a single unknown, m(arc)EHL, so we solve for it.
2m(arc)EHL = 3[360 - m(arc)EHL - m(arc)EHL]
2m(arc)EHL = 1080 - 6m(arc)EHL
8m(arc)EHL = 1080
m(arc)EHL = 135
Substitute the arc measure just found in Equation 3.
m<EYL = (1/3)m(arc)EHL
m<EYL = (1/3)(135)
m<EYL = 45
5^-7 * 5^12
Because the base is the same, add the exponents.
5^ (-7+12)= 5^5
Final answer: 5^5, or 3125
