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3 years ago

The triangle below is equilateral. Find the length of side x in simplest radical form

Mathematics

1 answer:

Molodets [167]3 years ago

6 0

Answer:

4sqrt(3)

Step-by-step explanation:

If the triangle is equilateral, the hypotenuse is 2x

We can use the Pythagorean theorem

a^2 + b^2 = c^2 where a and b are the legs and c is the hypotenuse

x^2 + 12^2 = (2x)^2

x^2 + 144 = 4x^2

144 = 4x^2 -x^2

144 = 3x^2

48 = x^2

Taking the square root of each side

sqrt(48) = sqrt(x^2)

sqrt(16*3) = x

4 sqrt(3) =x

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The area of a triangle is 30 square centimeters and its base is 10 cm. What is the height in cm of the triangle

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Answer:

The height of the triangle is $6$ cm.

Step-by-step explanation:

----------------------------------------

The formula for solving the area of a triangle is $A=\frac{1}{2}bh$ where $b$ stands for base and $h$ stands for height.

Let's substitute 30 $cm^2$ for the area and 10 for the base.

$30=\frac{1}{2} (10)h$

-------------------->>>>

Now, let's solve to find the height.

$30=\frac{1}{2}(10)h$

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Divide both sides by 5.

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3 years ago

Express 2x+1/(x-2)(x²+1) as a partial fraction.

Oduvanchick [21]

Answer:

Partial fraction = 1/(x-2) - x/(x^2+1)

Step-by-step explanation:

Question:

Express 2x+1/(x-2)(x²+1) as a partial fraction.

Note: it will be assumed that there was a typo in the interpretation of parentheses to mean

(2x+1) / ( (x-2)(x^2+1) )

Let

(2x+1) / ( (x-2)(x^2+1) ) = A/(x-2) + (Bx+C)/(x^2+1) .........................(0)

(2x+1) / ( (x-2)(x^2+1) ) = (A(x^2+1)+(Bx+C)(x-2)) / ( (x-2)(B/(x^2+1) )

(2x+1) / ( (x-2)(x^2+1) ) = (Ax^2+A+Bx^2+(C-2B)x-2C) / ( (x-2)(B/(x^2+1) )

(2x+1) / ( (x-2)(x^2+1) ) = ( (A+B)x^2+(C-2B)x+A-2C ) / ( (x-2)(B/(x^2+1) )

Match numerators

2x+1 = (A+B)x^2+(C-2B)x+A-2C

Match coefficients,

A+B = 0 ..................(1)

-2B+C = 2 .................(2)

A-2C = 1 ...................(3)

Solve for A, B and C

Substitute A from (1) in (3)

-B - 2C =1

transpose and solve for B

B = -2C-1 ....................(4)

Substitue B from (4) in (2)

-2(-2C-1) + C = 2

simplify

5C = 2-2 = 0

C=0 ..........................(5)

substitute (5) in (4)

B = -2C-1 = -1 ...............(6)

Substitue (6) in (1)

A+(-1) = 0

A=1 .............................(7)

Using values from (7), (6) and (5) to substitute in (0)

we get

(2x+1) / ( (x-2)(x^2+1) ) = 1/(x-2) - x/(x^2+1)

as the required partial fraction

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