All categories

3 years ago

In class, joe notices a triangular shaped desk. leg a is 4 and leg b is 10. find c.

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

7 0

A^2 + b^2 = c^2 fill in the know numbers a=4 b=10
$^2 +10^2 = 116
now find the square root of that its
c is approx 10.8

You might be interested in

How to convert the repeating nonterminating decimal to a fraction 0.1515

Ipatiy [6.2K]

The answer to the question

5 0

3 years ago

Find the value of x. <br> m27 = 2x + 18 and 2B

Reil [10]

8 0

2 years ago

Please give me the correct answer

vlada-n [284]

The exponent should be 6

3 0

3 years ago

Read 2 more answers

What is the result of adding these two equations?

timofeeve [1]

Answer:

the result of adding these two equations is

3x = 14

Step-by-step explanation:

We are given two equations

5x − y = 6 eq. 1

−2x + y = 8 eq. 2

Add the like terms together

5x - 2x = 3x

− y + y = 0

6 + 8 = 14

So the result of adding these equations is

3x + 0 = 14

3x = 14

Bonus:

The value of x is

3x = 14

x = 14/3

6 0

3 years ago

Find the general solution of x'1 = 3x1 - x2 + et, x'2 = x1.

Ann [662]

Note that if ${x_2}'=x_1$ , then ${x_2}''={x_1}'$ , and so we can collapse the system of ODEs into a linear ODE:

${x_2}''=3{x_2}'-x_2+e^t$
${x_2}''-3{x_2}'+x_2=e^t$

which is a pretty standard linear ODE with constant coefficients. We have characteristic equation

$r^2-3r+1=\left(r-\dfrac{3+\sqrt5}2\right)\left(r+\dfrac{3+\sqrt5}2\right)=0$

so that the characteristic solution is

${x_2}_C=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}$

Now let's suppose the particular solution is ${x_2}_p=ae^t$ . Then

${x_2}_p={{x_2}_p}'={{x_2}_p}''=ae^t$

and so

$ae^t-3ae^t+ae^t=-ae^t=e^t\implies a=-1$

Thus the general solution for $x_2$ is

$x_2=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}-e^t$

and you can find the solution $x_1$ by simply differentiating $x_2$ .

7 0

3 years ago