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3 years ago

What is the area of the triangular park in square meters ?

Mathematics

1 answer:

qaws [65]3 years ago

3 0

Answer:

A = 9 sq. meters

Step-by-step explanation:

From the graph, we can tell that it is a isosceles triange. So,

$A=\frac{1}{2}bh\\ A=\frac{1}{2}(6)(3)\\ A=9$

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2) Find the perimeter of the figure below.<br> 6yd 5yd 9yd 4yd

miv72 [106K]

You must add all sides together to get the perimeter.
The perimeter is 24

3 0

3 years ago

If two sides of a triangle are congruent, then the angles opposite those sides are _______ congruent.

madam [21]

Answer:

Always

Step-by-step explanation:

Properties of an Isoceles Triangle

6 0

3 years ago

A maintenance technician sights the top of a telephone pole at a 25° angle of elevation as shown. Technician Telephone Pole Dete

Bumek [7]

Answer:

$x = 375.3$

Step-by-step explanation:

Given

See attachment for illustration

Required

Determine the horizontal distance

The horizontal distance is the distance between the technician and the base of the pole.

Represent this with x.

The relationship between x, 175ft and 25 degrees is represented as:

$tan(25^{\circ}) = \frac{175}{x}$ --- i.e. tan formula

Multiply both sides by x

$x * tan(25^{\circ}) = \frac{175}{x} * x$

$x * tan(25^{\circ}) = 175$

Make x the subject

$x = \frac{175}{tan(25^{\circ})}$

$x = \frac{175}{0.4663}$

$x = 375.294874544$

$x = 375.3$ --- approximated

<em>The distance is approximately 375.3ft</em>

7 0

3 years ago

Which situation results in a final value of zero?

amid [387]

B and D because 3-3 =0 and 5-5= 0.

3 0

3 years ago

Suppose that a certain college class contains 35 students. of these, 17 are juniors, 20 are social science majors, and 12 are ne

seraphim [82]

A) Suppose $J$ is the set of juniors and $S$ is the set of social science majors. There are 35 students in all, but 12 of them don't belong to either set, so $|J\cup S|=35-12=23$ . Then

$|J\cap S|=|J|+|S|-|J\cup S|=17+20-23=14$

So the probability that a random student is both a junior and social science major is $\mathbb P(J\cap S)=\dfrac{14}{35}$

b) We're looking for the probability $\mathbb P(S\mid J)$ . By definition, this would be

$\mathbb P(S\mid J)=\dfrac{\mathbb P(S\cap J)}{\mathbb P(J)}=\dfrac{\frac{14}{35}}{\frac{17}{35}}=\dfrac{14}{17}$

7 0

3 years ago