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Mrac [35]
3 years ago
13

Which pairs of triangles can be shown to be congruent using

Mathematics
1 answer:
Rom4ik [11]3 years ago
4 0
I can do the matho on Friday and Friday I will u would have abc and uvw
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In 1995, Orlando, Florida, was about 175,000. At the time, the population was growing at a rate of about 2000 per year. Write an
gladu [14]

Answer:

y= 2000 * x + 175000

Step-by-step explanation:

Given is Population is 175,000

Increase in rate = 2000/year

Let x be the population in 1995 which is C= 175,000

y= Population after per year

We know the formula for slope intercept form is

y= mx + C     ............... (i)

Now from the suppose values and given

m= 2000

Now putting values in equation (i) gives

y= 2000 * x + 175000

Here x is the No of years

6 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
Rewrite the quadratic equation in standard form y=5(x-2)^2+7
shusha [124]

Answer:

To write a polynomial in standard form, simplify and then arrange the terms in descending order.

y=ax²+bx+c Simplify each term.

y=5x²−20x+20+7

Add 20 and 7.

y=5x²−20x+27

by-step explanation:

8 0
3 years ago
Read 2 more answers
A lab technician needs 35 ml of 15% base solution for a certain experiment,
kirza4 [7]

Answer:

17.5ml- of 10 percent solution, 17.5ml- of 20 percent solution

Step-by-step explanation:

35:100*15=5.25- ml of alkali in the base solution

Suppose we need x ml of 10 percents solution and 35-x - of 20 percents.

Then The quantity of alkali in the first one (10 percents) is x/100*10=0.1x

when in the second one we have (35-x)/100*20= 7-0.2x of alkali

0.1x+7-0.2x=5.25

7-0.1x= 5.25

0.1x=1.75

x=17.5- 0f 10 percents

35-17.5=17.5 - of 20 percents

5 0
3 years ago
the temperature in minnesota was -8°F on day. On the same day, temperature was 12° higher in India . What was the temperature in
stellarik [79]

it would be a positive 4 degrees F today

6 0
3 years ago
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