All categories

3 years ago

∠2 and ∠3 form a linear pair. ∠3 and ∠1 are vertical angles. m∠1= 62 ∘ . What is m∠3?

2 answers:

8 0

For this problem, we do not need the first portion of the given information, so forget it for now. All we need to know is that 3 and 1 are vertical angles. Verticle angles are equal.
1=62
3 = 1
3 = 62
The measure of angle 3 is 62 degrees

Hope this helps =)

Rashid [163]3 years ago

3 0

M<3 is he sam as 123

You might be interested in

If X is a r.v. such that E(X^n)=n! Find the m.g.f. of X,Mx(t). Also find the ch.f. of X,and from this deduce the distribution of

astraxan [27]

$M_X(t)=\mathbb E(e^{Xt})$
$M_X(t)=\mathbb E\left(1+Xt+\dfrac{t^2}{2!}X^2+\dfrac{t^3}{3!}X^3+\cdots\right)$
$M_X(t)=\mathbb E(1)+t\mathbb E(X)+\dfrac{t^2}{2!}\mathbb E(X^2)+\dfrac{t^3}{3!}\mathbb E(X^3)+\cdots$
$M_X(t)=1+t+t^2+t^3+\cdots$
$M_X(t)=\displaystyle\sum_{k\ge0}t^k=\frac1{1-t}$

provided that $|t|$ .

Similarly,

$\varphi_X(t)=\mathbb E(e^{iXt})$
$\varphi_X(t)=1+it+(it)^2+(it)^3+\cdots$
$\varphi_X(t)=(1-t^2+t^4-t^6+\cdots)+it(1-t^2+t^4-t^6+\cdots)$
$\varphi_X(t)=(1+it)(1-t^2+t^4-t^6+\cdots)$
$\varphi_X(t)=\dfrac{1+it}{1+t^2}=\dfrac1{1-it}$

You can find the CDF/PDF using any of the various inversion formulas. One way would be to compute

$F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{e^{itx}\varphi_X(-t)-e^{-itx}\varphi_X(t)}{it}\,\mathrm dt$

The integral can be rewritten as

$\displaystyle\int_0^\infty\frac{2i\sin(tx)-2it\cos(tx)}{it(1+t^2)}\,\mathrm dt$

so that

$F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt$

There are lots of ways to compute this integral. For instance, you can take the Laplace transform with respect to $x$ , which gives

$\displaystyle\mathcal L_s\left\{\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt\right\}=\int_0^\infty\frac{1-s}{(1+t^2)(s^2+t^2)}\,\mathrm dt$
$=\displaystyle\frac{\pi(1-s)}{2s(1+s)}$

and taking the inverse transform returns

$F_X(x)=\dfrac12+\dfrac1\pi\left(\dfrac\pi2-\pi e^{-x}\right)=1-e^{-x}$

which describes an exponential distribution with parameter $\lambda=1$ .

6 0

3 years ago

Given the function rule f(x)=x2-4x+3 what is the output of f(-3)

Mariana [72]

(-3)^2 -4(-3) + 3

9 + 12 + 3

So the answer is 24

3 0

3 years ago

If cos A = 3 over 11, then which of the following is correct?

ziro4ka [17]

Answer: The answer is (a) sec A = 11 over 3.

Step-by-step explanation: Given that Cosine of an angle 'A' is 3 over 11,

i.e.,

$\cos A=\dfrac{3}{11}.$

And we need to find which one of the given four options is correct.

We have the following relations between cosine, secant and cosecant of an angle from trigonometry.

$\sec A=\dfrac{1}{\cos A}~~\textup{and}~~\csc A=\dfrac{1}{\sqrt{1-\cos^2 A}}.$

Therefore,

$\sec A=\dfrac{11}{3}$

and

$\csc A=\dfrac{1}{\sqrt{1-\frac{9}{121}}}=\dfrac{1}{\sqrt{\frac{112}{121}}}=\dfrac{11}{4\sqrt 7}.$

Thus, the correct option is (a) sec A = 11 over 3.

3 0

3 years ago

A printer takes 30 minutes to print a banner and a poster. I will take 50 minutes to print a banner and three posters. How long

qwelly [4]

Answer:

It takes the printer 20 minutes to print a banner

Step-by-step explanation:

Notice that there are two unknowns in this problem: the time to print a poster and the time to print a banner. Let's assign the letter "b" to the time it takes to print a banner, and "p' the time it takes to print a poster.

So we can write the following two equations involving the total amount of time that takes to print in the two cases:

$b+p=30\\b+3p=50$

We can solve for the unknown "p" by subtracting term by term in the equation set above:

$b+3p=50\\b+p=30\\\\b-b+3p-p=50-30\\2p=20\\p=10$

So it takes 10 minutes to print each poster, we can find the time it takes to print the banner by using the value we just found in one of the equations:

$b+p=30\\b+10=30\\b=20$

So, it takes 20 minutes to print a banner.

8 0

3 years ago

JOJO help me I can't work good

photoshop1234 [79]

Answer:

C. 46 degrees

Step-by-step explanation:

In a rotation all angles and side lengths are preserved so the angle measure will be the same in both the image and pre-image.

6 0

3 years ago

Read 2 more answers