All categories

3 years ago

The m<3 = (2x + 10) and m<6 = (3x-15) find the value of x, m<3 and M<6

Mathematics

1 answer:

sergeinik [125]3 years ago

4 0

9514 1404 393

Answer:

x = 25; ∠3 = ∠6 = 60°

Step-by-step explanation:

Angles 3 and 6 are alternate interior angles, so are congruent.

2x +10 = 3x -15

25 = x . . . . . . . . . . . add 15-2x

2x +10 = 2(25) +10 = 60

The measures of angles 3 and 6 are 60°.

You might be interested in

A 6-pound bag of sugar contains 681 one-teaspoon servings and costs $4.49. A batch of muffins uses 1 cup of sugar. How many whol

irina1246 [14]

681 tsp/48 tsp = 14.1875
If you round down, you get 14 servings.

6 0

3 years ago

Divide. Write the quotient in lowest terms.

const2013 [10]

Answer:

$\frac{11}{12}$

7 0

3 years ago

How do I solve 2x-3y=12 step by step?

anzhelika [568]

Answer:

Step-by-step explanation: •Subtract 2x from both sides of the equation.

-3y = 12 - 2x

•Divide each term by -3y = 12 - 2x by -3.

-3y/-3 = 12/-3 + -2x/-3

•Cancel the common factor of -3.

12/-3 + -2x/-3

•Divide y by 1.

y = 12/-3 + -2x/-3

•Simplify each term.

•Divide 12 by -3.

y = -4 + -2x/-3

•Dividing two negatives values results in a positive value.

y = -4 + 2x/3

4 0

3 years ago

Find the probability of rolling the difference of 1

Naddik [55]

Answer:

It’s 1

Step-by-step explanation:

7 0

4 years ago

Find a particular solution to the differential equation y′′+2y′+y=2t2−t+3e−2t. y′′+2y′+y=2t2−t+3e−2t.

Jlenok [28]

$y''+2y'+y=2t^2-t+3e^{-2t}$

The characteristic equation is

$r^2+2r+1=(r+1)^2=0\implies r=-1$

so the characteristic solution is

$y_c=C_1e^{-t}+C_2te^{-t}$

For the particular solution, we can try looking for a solution of the form

$y_p=at^2+bt+c+de^{-2t}$
$\implies{y_p}'=2at+b-2de^{-2t}$
$\implies{y_p}''=2a+4de^{-2t}$

Substituting into the ODE, we have

$(2a+4de^{-2t})+2(2at+b-2de^{-2t})+(at^2+bt+c+de^{-2t})=2t^2-t+3e^{-2t}$
$at^2+(4a+b)t+(2a+2b+c)+de^{-2t}=2t^2-t+3e^{-2t}$
$\implies\begin{cases}a=2\\4a+b=-1\\2a+2b+c=0\\d=3\end{cases}\implies a=2,b=-9,c=14,d=3$

So the general solution to the ODE is

$y=y_c+y_p$
$y=C_1e^{-t}+C_2te^{-t}+2t^2-9t+14+3e^{-2t}$

4 0

3 years ago