All categories

3 years ago

The length of the hypotenuse of 30-60-90 triangle is 5. find the perimeter

Mathematics

1 answer:

baherus [9]3 years ago

4 0

Answer:

$7.5+2.5\sqrt{3}$

Step-by-step explanation:

The ratio of sides in a 30-60-90 triangle is $1:2:\sqrt{3}$ . Since the hypotenuse is the longest side, the shortest leg is 5/2=2.5. The leg that would usually be $\sqrt{3}$ is now $2.5\sqrt{3}\approx 4.33$ . Adding all of these together, you get $7.5+2.5\sqrt{3}\approx 11.83$ . Hope this helps!

You might be interested in

Four times the largest of three consecutive odd integers is 36 more than the sum of the other

erica [24]

If you know how to solve word problems involving the sum of consecutive even integers, you should be able to easily solve word problems that involve the sum of consecutive odd integers. The key is to have a good grasp of what odd integers are and how consecutive odd integers can be represented.

Odd Integers

If you recall, an even integer is always 22 times a number. Thus, the general form of an even number is n=2kn=2k, where kk is an integer.

So what does it mean when we say that an integer is odd? Well, it means that it’s one less or one more than an even number. In other words, odd integers are one unit less or one unit more of an even number.

Therefore, the general form of an odd integer can be expressed as nn is n=2k-1n=2k−1 or n=2k+1n=2k+1, where kk is an integer.

Observe that if you’re given an even integer, that even integer is always in between two odd integers. For instance, the even integer 44 is between 33 and 55.

6 0

3 years ago

The perimeter of a playing field for a certain sport is 254 ft. The length is 47 ft longer than the width. Find the dimensions.

Vikentia [17]

Answer: the length is 87 feet

The width is 40 feet

Step-by-step explanation:

Let L represent the length of the playing field.

Let W represent the width of the playing field.

The playing field is rectangular. The formula for determining the perimeter of a rectangle is expressed as

Perimeter = 2(L + W)

The perimeter of a playing field for a certain sport is 254 ft. This means that

254 = 2(L + W)

L + W = 254/2

L + W = 127 - - - - - - - - - - - -1

The length is 47 ft longer than the width. This means that

L = W + 47

Substituting L = W + 47 into equation 1, it becomes

W + 47 + W = 127

2W + 47 = 127

2W = 127 - 47 = 80

W = 80/2 = 40

L = W + 47 = 40 + 47

L = 87

4 0

3 years ago

HELP ASAP! TY PLEASE ALSO TRY TO EXPLAIN

Bas_tet [7]

It’s b :) #girlboss #gatekeep #gaslight #uwu

4 0

2 years ago

Explain how you decided what number sentence to write for exercise 1 how to do that

Musya8 [376]

The answer is 21 i really don't know

8 0

3 years ago

Consider the differential equation y'' − y' − 20y = 0. Verify that the functions e−4x and e5x form a fundamental set of solution

KIM [24]

Answer:

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

Step-by-step explanation:

Given differential equation is

y''-y'-20y =0

Here P(x)= -1, Q(x)= -20 and R(x)=0

Let trial solution be $y=e^{mx}$

Then, $y'=me^{mx}$ and $y''=m^2e^{mx}$

$\therefore m^2e^{mx}-m e^{mx}-20e^{mx}=0$

$\Rightarrow m^2-m-20=0$

$\Rightarrow m^2-5m+4m-20=0$

$\Rightarrow m(m-5)+4(m-5)=0$

$\Rightarrow (m-5)(m+4)=0$

$\Rightarrow m=-4,5$

Therefore the complementary function is = $c_1e^{-4x}+c_2e^{5x}$

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

If $y_1$ and $y_2$ are the fundamental solution of differential equation, then

$W(y_1,y_2)=\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|\neq 0$

Then $y_1$ and $y_2$ are linearly independent.

$W(e^{-4x},e^{5x})=\left|\begin{array}{cc}e^{-4x}&e^{5x}\\-4e^{-4x}&5e^{5x}\end{array}\right|$

$=e^{-4x}.5e^{5x}-e^{5x}.(-4e^{-4x})$

$=5e^x+4e^x$

$=9e^x\neq 0$

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

Let the the particular solution of the differential equation is

$y_p=v_1e^{-4x}+v_2e^{5x}$

$\therefore v_1=\int \frac{-y_2R(x)}{W(y_1,y_2)} dx$

and

$\therefore v_2=\int \frac{y_1R(x)}{W(y_1,y_2)} dx$

Here $y_1= e^{-4x}$ , $y_2=e^{5x}$ , $W(e^{-4x},e^{5x})=9e^x$ ,and $R(x)=0$

$\therefore v_1=\int \frac{-e^{5x}.0}{9e^x}dx$

and

$\therefore v_2=\int \frac{e^{5x}.0}{9e^x}dx$

The the P.I = 0

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

7 0

3 years ago