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Georgia [21]
3 years ago
10

What is the slope of the line that contains these points

Mathematics
1 answer:
abruzzese [7]3 years ago
7 0

Answer:

Hi there!

The slope of the line is: 1/5

Step-by-step explanation:

the slope formula is delta y over delta x

in other words: (Y-Y1)/ (X-X1)

so you are free to choose any two points you want in this case I chose to use (11,0) and (6,-1) so plugging it in the equation you get: (0-(-1))/(11-6)

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I think it 2/5 or -2/5
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At 20 mph the force of your car impacting a surface is about four times as great as 10 mph.
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Read 2 more answers
The third term of an arithmetical progression is 7, and the seventh term is 2 more than 3 times the third term.Find the first te
Sphinxa [80]
General formula for n-th term of arithmetical progression is

a(n)=a(1)+d(n-1).

For 3d term we have

a(3)=a(1) +d(3-1),  where a(3)=7
7=a(1)+2d

For 7th term we have
a(7)=a(1) +d(7-1)
a(7)=a(1) + 6d

Also, we have that the <span>seventh term is 2 more than 3 times the third term,
a(7)=3*a(3)+2= 3*7+2=21+2=23

So we have, </span>a(7)=a(1) + 6d and a(7)=23. We can write
23=a(1) + 6d.

Now we can write a system of equations
23=a(1) + 6d
<span> - (7=a(1)+2d)
</span>
16 = 4d
d=4,

7=a(1)+2d

7=a(1)+2*4
a(1)=7-8=-1
a(1)= - 1

First term a(1)=-1, common difference d=4.

Sum of  the 20 first terms is
S=20 * (a(1)+a(20))/2

a(1)=-1
a(n)=a(1)+d(n-1)
a(20) = -1+4(20-1)=-1+4*19=75

S=20 * (-1+75)/2=74*10=740

Sum of 20 first terms is 740.
3 0
3 years ago
Find the cube roots of 27(cos 330° + i sin 330°)
Aleksandr-060686 [28]

Answer:

See below for all the cube roots

Step-by-step explanation:

<u>DeMoivre's Theorem</u>

Let z=r(cos\theta+isin\theta) be a complex number in polar form, where n is an integer and n\geq1. If z^n=r^n(cos\theta+isin\theta)^n, then z^n=r^n(cos(n\theta)+isin(n\theta)).

<u>Nth Root of a Complex Number</u>

If n is any positive integer, the nth roots of z=rcis\theta are given by \sqrt[n]{rcis\theta}=(rcis\theta)^{\frac{1}{n}} where the nth roots are found with the formulas:

  • \sqrt[n]{r}\biggr[cis(\frac{\theta+360^\circ k}{n})\biggr] for degrees (the one applicable to this problem)
  • \sqrt[n]{r}\biggr[cis(\frac{\theta+2\pi k}{n})\biggr] for radians

for  k=0,1,2,...\:,n-1

<u>Calculation</u>

<u />z=27(cos330^\circ+isin330^\circ)\\\\\sqrt[3]{z} =\sqrt[3]{27(cos330^\circ+isin330^\circ)}\\\\z^{\frac{1}{3}} =(27(cos330^\circ+isin330^\circ))^{\frac{1}{3}}\\\\z^{\frac{1}{3}} =27^{\frac{1}{3}}(cos(\frac{1}{3}\cdot330^\circ)+isin(\frac{1}{3}\cdot330^\circ))\\\\z^{\frac{1}{3}} =3(cos110^\circ+isin110^\circ)

<u>First cube root where k=2</u>

<u />\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(2)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+720^\circ}{3})\biggr]\\3\biggr[cis(\frac{1050^\circ}{3})\biggr]\\3\biggr[cis(350^\circ)\biggr]\\3\biggr[cos(350^\circ)+isin(350^\circ)\biggr]

<u>Second cube root where k=1</u>

\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(1)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+360^\circ}{3})\biggr]\\3\biggr[cis(\frac{690^\circ}{3})\biggr]\\3\biggr[cis(230^\circ)\biggr]\\3\biggr[cos(230^\circ)+isin(230^\circ)\biggr]

<u>Third cube root where k=0</u>

<u />\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(0)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ}{3})\biggr]\\3\biggr[cis(110^\circ)\biggr]\\3\biggr[cos(110^\circ)+isin(110^\circ)\biggr]

4 0
2 years ago
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