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3 years ago

What is a 200yd football field in meters in Europe

Mathematics

1 answer:

timofeeve [1]3 years ago

8 0

Answer:

180 meters

Step-by-step explanation:

100yd field is 90 meters so a 200yd field is 180 meters

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Find the x- and y-intercepts of the equation below. Write your answers as ordered pairs.

Airida [17]

Answer:

x intercept: none

y intercept=: (0,-12)

5 0

3 years ago

an umbrella is made by stitching 10 triangular pieces of cloth of two different colours where each pieces measuring 48cm,48cm,an

mars1129 [50]

Answer:

Step-by-step explanation:

first we need to calculate the area of isosceles triangle having sides 48cm, 48cm and 30cm

a = 48

b= 30

Area of isosceles triangle using only sides = ½[√(a^2 − b^2 /4) × b]

= 1/2*30√( 48² - 1/4×30²)

=1/2*30√ (2304 -225)

=684sq cm

cost required to make umbrella= 10*684*5

= 3<em>4200</em>

7 0

3 years ago

In order to answer the question correctly, please use the following image below:

Maru [420]

Answer:

see below

Step-by-step explanation:

The angle where chords cross is the average of the intercepted arcs. Here, that is ...

(37° +46°)/2 = (83°)/2 = 41.5°

Angle SUT is 41.5°.

_____

<em>Comment on the error</em>

The measure of an arc cannot be arbitrarily said to be the same as the angle where the chords cross. It will be the same if (a) the chords cross at the circle center, or (b) the opposite intercepted arc has the same measure. Neither of these conditions hold here.

8 0

3 years ago

Read 2 more answers

Use the Ratio Test to determine the convergence or divergence of the series. If the Ratio Test is inconclusive, determine the co

jeka57 [31]

Answer:

<h2>A. The series CONVERGES</h2>

Step-by-step explanation:

If $\sum a_n$ is a series, for the series to converge/diverge according to ratio test, the following conditions must be met.

$\lim_{n \to \infty} |\frac{a_n_+_1}{a_n}| = \rho$

If $\rho$ < 1, the series converges absolutely

If $\rho > 1$ , the series diverges

If $\rho = 1$ , the test fails.

Given the series $\sum\left\ {\infty} \atop {1} \right \frac{n^2}{5^n}$

To test for convergence or divergence using ratio test, we will use the condition above.

$a_n = \frac{n^2}{5^n} \\a_n_+_1 = \frac{(n+1)^2}{5^{n+1}}$

$\frac{a_n_+_1}{a_n} = \frac{{\frac{(n+1)^2}{5^{n+1}}}}{\frac{n^2}{5^n} }\\\\ \frac{a_n_+_1}{a_n} = {{\frac{(n+1)^2}{5^{n+1}} * \frac{5^n}{n^2}\$

$\frac{a_n_+_1}{a_n} = {{\frac{(n^2+2n+1)}{5^n*5^1}} * \frac{5^n}{n^2}\\$

aₙ₊₁/aₙ =

$\lim_{n \to \infty} |\frac{ n^2+2n+1}{5n^2}| \\\\Dividing\ through\ by \ n^2\\\\\lim_{n \to \infty} |\frac{ n^2/n^2+2n/n^2+1/n^2}{5n^2/n^2}|\\\\\lim_{n \to \infty} |\frac{1+2/n+1/n^2}{5}|\\\\$