All categories

3 years ago

I need help , if yk the answer help fr!!!

Mathematics

1 answer:

Feliz [49]3 years ago

4 0

Answer:

The measure of the angle JKG is:

m∠JKG = 56°

Step-by-step explanation:

<u>Given</u>

m∠JKG = 76-2x

m∠FHK = 6x-4

J is a midpoint of the segment FG and K is a midpoint of the segment GH.

<u>To determine</u>

m∠JKG = ?

Given that J is a midpoint of the segment FG and K is a midpoint of the segment GH. Thus, making two similar triangles, ΔJGK and ΔFGH

We know that two triangles are similar if the only difference is size. So, the angles remain the same.

so m∠JKG and m∠FHK are equal.

i.e.

m∠JKG = m∠FHK

substitute m∠JKG = 76-2x and m∠FHK = 6x-4

76-2x = 6x-4

6x+2x = 76 + 4

8x = 80

divide both sides by 8

8x/8 = 80/8

x = 10

Therefore, the value of x = 10

m∠JKG = 76-2x

substitute x = 10

m∠JKG = 76 - 2(10)

= 76 - 20

= 56°

Therefore, measure of the angle JKG is:

m∠JKG = 56°

You might be interested in

Jerry best with 40 questions, out of which he answered 22 correctly. To pass the test, he needs at least 60% correct. What was h

a_sh-v [17]

Answer:

22?

Step-by-step explanation:

I sorry if my answer is wrong.

4 0

3 years ago

Read 2 more answers

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

A train travels 95 kilometers in 4 hours, and then 94 kilometers in 3 hours. What is its average speed?

Brilliant_brown [7]

One way you could do this is by dividing 94 by 3 and then 95 by 4. This gives you the KPH (Kilometers per hour) of both trains. The first speed you mentioned, we will call that speed 'A.' The second one will be speed 'B.' Speed A's KPH is 23.75 MPH. Speed B's KPH is 31.333. We can change these into fractions (23 3/4 and 31 1/3.) These fractions could then be set into like terms. (This gives us 285/12 and 376/12) We then add these together. This gives us 661/12. The next thing to do is to convert this back to a decimal form, which gives us 55.0833.... Divide this number by two, and that is your answer. The average speed of the train is 29.04 KPH (This number is rounded to the nearest hundreth- be sure to make a note of that on your answer sheet)

5 0

3 years ago

2%exchange price and the bass price is 160,000 how much in total

Ludmilka [50]

2% of 160000
(2/100) * 160000
3200

Total amount
160000 + 3200
163200

7 0

3 years ago

If ∠A and ∠B are vertical angles, and ∠A = 66°, find ∠B.

Vikentia [17]

By definition, vertical angles are equal. so Angle B = Angle A = 66 degrees

7 0

4 years ago

Read 2 more answers