All categories

2 years ago

A circular pond has a radius of 35 feet. What is the diameter of the pond?

Mathematics

1 answer:

SVEN [57.7K]2 years ago

4 0

Answer:

I believe the radius is half the circle which makes the diameter 2x the radius which is 35+35= 70

You might be interested in

I need help with this one question.

Tom [10]

Answer:

answer a

Step-by-step explanation:

3 0

3 years ago

In AOPQ, o = 700 cm, p = 840 cm and q=620 cm. Find the measure of _P to the<br> nearest degree

Westkost [7]

Given:

In triangle OPQ, o = 700 cm, p = 840 cm and q=620 cm.

To find:

The measure of angle P.

Solution:

According to the Law of Cosines:

$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$

Using Law of Cosines in triangle OPQ, we get

$\cos P=\dfrac{o^2+q^2-p^2}{2oq}$

$\cos P=\dfrac{(700)^2+(620)^2-(840)^2}{2(700)(620)}$

$\cos P=\dfrac{490000+384400-705600}{868000}$

$\cos P=\dfrac{168800}{868000}$

On further simplification, we get

$\cos P=0.19447$

$P=\cos^{-1}(0.19447)$

$P=78.786236$

$P\approx 79$

Therefore, the measure of angle P is 79 degrees.

8 0

3 years ago

Read 2 more answers

Yoo how yall doinnnnn

ICE Princess25 [194]

Answer:

pretty good tired of the virus stuff but good

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Will give brainliest answer

yaroslaw [1]

Answer

Step-by-step explanation:

To find the area of a shape multiply its height by its width. For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself to find the area.

please mark me as brainliest.

7 0

3 years ago

Read 2 more answers

Which would be the best method to use to solve the following equations? Explain your reasoning.

o-na [289]

A. square root property

$3x^2-192=0\\x^2-66=0\\x^2=66\\x= -\sqrt{66} or \sqrt{66}$

it has one value with x which is x^2 and it cn be easily solved without having to factorise, quadratic formula cant be used as it need ax^2+bx+c=0 format

B. factorising

$x^2-x-6=0\\x^2+2x-3x-6=0\\(x+2)(x-3)=0\\x+2=0 and x-3=0\\x= -2 \\x= 3$

i just felt like this was easier to factorise than the other 2 options left

C. Completing the square

$x^2-6x-7=0\\(x-3)^2=0\\x= -1\\x= 7$

same reason personal preference

D- Quadratic

$x^2-17x-7=0\\x=\frac{-(-17)+\sqrt{(-17)^2-4(1)(-7)} }{2(1)} \\x=\frac{-(-17)-\sqrt{(-17)^2-4(1)(-7)} }{2(1)}\\x=17.4\\x=-0.402$

the 17 kinda threw me off and i didnt wanna get on factorising or doing completing the square so quadratic formal

7 0

2 years ago