All categories

3 years ago

Please help gotta pass dis

Mathematics

2 answers:

coldgirl [10]3 years ago

7 0

Hi there, so we know that all triangles add up to 180 degrees so we will take 38 and add it with 25 to get 63 then we subtract 180 by 63 to get 117 which is your awnser!! I mean I might be wrong but I’m probably right. I hope this helps!! Have a great morning, night, or day, Whatever it is for you!

Jlenok [28]3 years ago

4 0

Answer:

d=63 degree (if one side of a triangle is produced,the exterior angle so formed is equal to the sum of the two interior opposite angles).

Step-by-step explanation:

Hope it helps you!!

You might be interested in

Give the coordinates of a point on the line whose equation in point-slope form is y − (−3) = 1 4 (x − 9).

statuscvo [17]

Answer:

Below

Step-by-step explanation:

● y -(-3) = (1/4) (x-9)

Replace 1/4 by 0.25 to make it easier

● y + 3 = 0.25(x-9)

● y + 3 = 0.25x - 2.25

Add 3 to both sides

● y +3-3 = 0.25x -2.25-3

● y = 0.25x -5.25

To the coordinates of a point from this line replace x by a value.

The easiest one is 0.

● y = 0.25×0 -5.25

● y = 5.25

5.25 is 21/4

So the coordinates are (0, 21/4)

7 0

3 years ago

A golfer's score after playing on Friday was +2. His score for Saturday's round was -5. At the end of his round on Sunday, he wa

Lesechka [4]

The correct answer is-2

3 0

3 years ago

Read 2 more answers

For a standard normal distribution, if P (z less-than-or-equal-to a) = 0.7116, what is the value of P (z greater-than-or-equal-t

Irina-Kira [14]

Answer:equal

Step-by-step explanation:

The probability that a random variable is greater than or equal to z standard deviations from the mean in a standard normal distribution he P(z≤a)+P(z>a)=1, so P(z>a)=1-P(z≤a)=1-.7116=.2884

6 0

2 years ago

Plz help me <br>who can answer in just 1 min

nexus9112 [7]

9×9=81 so it would be 9cm

8 0

3 years ago

Compute the sum:

Nady [450]

You could use perturbation method to calculate this sum. Let's start from:

$S_n=\sum\limits_{k=0}^nk!\\\\\\\(1)\qquad\boxed{S_{n+1}=S_n+(n+1)!}$

On the other hand, we have:

$S_{n+1}=\sum\limits_{k=0}^{n+1}k!=0!+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=0}^{n}(k+1)!=\\\\\\=1+\sum\limits_{k=0}^{n}k!(k+1)=1+\sum\limits_{k=0}^{n}(k\cdot k!+k!)=1+\sum\limits_{k=0}^{n}k\cdot k!+\sum\limits_{k=0}^{n}k!\\\\\\(2)\qquad \boxed{S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n}$

So from (1) and (2) we have:

$\begin{cases}S_{n+1}=S_n+(n+1)!\\\\S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\end{cases}\\\\\\
S_n+(n+1)!=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\\\\\\
(\star)\qquad\boxed{\sum\limits_{k=0}^{n}k\cdot k!=(n+1)!-1}$

Now, let's try to calculate sum $\sum\limits_{k=0}^{n}k\cdot k!$ , but this time we use perturbation method.

$S_n=\sum\limits_{k=0}^nk\cdot k!\\\\\\
\boxed{S_{n+1}=S_n+(n+1)(n+1)!}\\\\\\
$

but:

$S_{n+1}=\sum\limits_{k=0}^{n+1}k\cdot k!=0\cdot0!+\sum\limits_{k=1}^{n+1}k\cdot k!=0+\sum\limits_{k=0}^{n}(k+1)(k+1)!=\\\\\\=
\sum\limits_{k=0}^{n}(k+1)(k+1)k!=\sum\limits_{k=0}^{n}(k^2+2k+1)k!=\\\\\\=
\sum\limits_{k=0}^{n}\left[(k^2+1)k!+2k\cdot k!\right]=\sum\limits_{k=0}^{n}(k^2+1)k!+\sum\limits_{k=0}^n2k\cdot k!=\\\\\\=\sum\limits_{k=0}^{n}(k^2+1)k!+2\sum\limits_{k=0}^nk\cdot k!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\
\boxed{S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n}$

When we join both equation there will be:

$\begin{cases}S_{n+1}=S_n+(n+1)(n+1)!\\\\S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\end{cases}\\\\\\
S_n+(n+1)(n+1)!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\\\
\sum\limits_{k=0}^{n}(k^2+1)k!=S_n-2S_n+(n+1)(n+1)!=(n+1)(n+1)!-S_n=\\\\\\=
(n+1)(n+1)!-\sum\limits_{k=0}^nk\cdot k!\stackrel{(\star)}{=}(n+1)(n+1)!-[(n+1)!-1]=\\\\\\=(n+1)(n+1)!-(n+1)!+1=(n+1)!\cdot[n+1-1]+1=\\\\\\=
n(n+1)!+1$

So the answer is:

$\boxed{\sum\limits_{k=0}^{n}(1+k^2)k!=n(n+1)!+1}$

Sorry for my bad english, but i hope it won't be a big problem :)

8 0

4 years ago