<span>According to a map, the distance between Hilltop and Valleyfloor is 1 3/4 inches. If the map uses a scale of ¼ inch to represent 1 mile, what is the actual distance in miles from Hilltop to Valleyfloor?</span>
For this case we have:
a = 30 cm
c = 16 cm
We look for the length of the diagonal:
d = x + y
Where,
For x:
a ^ 2 = x ^ 2 + x ^ 2
x = a / root (2) = 30 / root (2) = 21.2132 cm
For y:
c ^ 2 = y ^ 2 + y ^ 2
y = c / root (2) = 16 / root (2) = 11.3137 cm
The diagonal is:
d = x + y
d = 21.2132 + 11.3137
d = 32.5269 cm
Then, the height is:
h = h1 + h2
For h1:
h1 = root (x ^ 2 - (a / 2) ^ 2) = root ((21.2132) ^ 2 - (30/2) ^ 2)
h1 = 15 cm
For h2:
h2 = root (y ^ 2 - (c / 2) ^ 2) = root ((11.3137) ^ 2 - (16/2) ^ 2)
h2 = 8 cm
Finally:
h = h1 + h2
h = 15 + 8
h = 23 cm
Then, the area is:
A = (1/2) * (a + c) * (h)
A = (1/2) * (30 + 16) * (23)
A = 529 cm ^ 2
Answer:
the area of an isosceles trapezoid is:
A = 529 cm ^ 2
Answer:
8
[2] +[-6]
2 + 6 = 8
Step-by-step explanation:
If c is a perfect square, then there is an even number of rectangular models that represent the polynomial. Then the correct option is B.
<h3>What is a polynomial?</h3>
Consider the polynomial x² + bx + c, where b and c are positive integers. If the polynomial is factorable, it can be modeled with a rectangle with a distinct length and width.
The table shows the relationship between some values of c and the number of rectangular models.
Then the relationship between c and the number of rectangular models representing a factorable polynomial x² + bx + c will be
If c is a perfect square, then there is an even number of rectangular models that represent the polynomial.
Then the correct option is B.
More about the polynomial link is given below.
brainly.com/question/17822016
#SPJ1
<h2>
Answer with explanation:-</h2>
Let be the population mean.
By considering the given information , we have
, since the alternative hypothesis is two tailed , so the test is two tail test.
Given : Sample size : n=30
Sample mean :
Then, Standard deviation :
The p-value = 0.1111, the null hypothesis is not rejected.
Thus , we conclude that we have evidence to accept the company's claim that the batteries have on average 500 charges.