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3 years ago

How do you find the hypotenuse

1 answer:

6 0

Given two right triangle legs
Use the Pythagorean theorem to calculate the hypotenuse from right triangle sides. Take a square root of sum of squares:
c = √(a² + b²)

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Explain what i did in a detailed answer.

forsale [732]

Answer:

Step-by-step explanation: You added the following numbers multiplied the product by 5 = 17 and then you abbrieviated it and reapeated the process

3 0

2 years ago

Five years from now, the sum of the ages of a women and her daughter will be 40 years. The difference in their present age is 24

Tomtit [17]

Answer:

Daughter age = 3 years

Step-by-step explanation:

Let x be the age of the women and y be the age of the daughter.

Given:

After five year the sum of the women and daughter age = 40

$(x+5)+(y+5)=40$

At present the sum of the women and daughter age

$x+5+y+5=40$

$x+y+10=40$

$x+y=40-10$

$x+y=30$ --------------(1)

So the sum of the present age is $x+y=30$

The difference in their present age is 24 years.

$x-y=24$

$x=24+y$

Now we substitute x value in equation 1.

$(24+y)+y=30$

$24+2y=30$

$2y=30-24$

$2y=6$

$y=\frac{6}{2}$

$y=3\ years$

Therefore, the daughter age is 3 years.

7 0

3 years ago

The line plot shows the length of rope (in feet) of five pieces of rope. What is the difference in length between the longest an

Vikki [24]

Answer: am sorry G i can't get that answers for you

Step-by-step explanation:

8 0

2 years ago

Sonia is planning a rectangular vegetable garden using a roll of border fencing that is 45 3/4 feet long. If she makes the width

Verizon [17]

Answer: 12 3/8 10 1/2 + 10 1/2= 21 45 3/4 - 21 = 24 3/4 24 ÷ 2 = 12 Change the 3/4 into 8ths = 6/8 Divide that in half which gives you 3/8 So the length is 12 3/8

5 0

2 years ago

Read 2 more answers

Simplify f+g / f-g when f(x)= x-4 / x+9 and g(x)= x-9 / x+4

steposvetlana [31]

$f(x)=\dfrac{x-4}{x+9};\ g(x)=\dfrac{x-9}{x+4}\\\\f(x)+g(x)=\dfrac{x-4}{x+9}+\dfrac{x-9}{x+4}=\dfrac{(x-4)(x+4)+(x-9)(x+9)}{(x+9)(x+4)}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2+x^2-9^2}{(x+9)(x+4)}=\dfrac{2x^2-16-81}{(x+9)(x+4)}=\dfrac{2x^2-97}{(x+9)(x+4)}\\\\f(x)-g(x)=\dfrac{x-4}{x+9}-\dfrac{x-9}{x+4}=\dfrac{(x-4)(x+4)-(x-9)(x+9)}{(x+9)(x+4)}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2-(x^2-9^2)}{(x+9)(x+4)}=\dfrac{x^2-16-x^2+81}{(x+9)(x+4)}=\dfrac{65}{(x+9)(x+4)}$

$\dfrac{f+g}{f-g}=(f+g):(f-g)=\dfrac{2x^2-97}{(x+9)(x+4)}:\dfrac{65}{(x+9)(x+4)}\\\\=\dfrac{2x^2-97}{(x+9)(x+4)}\cdot\dfrac{(x+9)(x+4)}{65}\\\\Answer:\ \boxed{\dfrac{f+g}{f-g}=\dfrac{2x^2-97}{65}}$

6 0

3 years ago

Read 2 more answers