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2 years ago

What is the equation of the line that passes through the points (5,3) and (9,-13)?

Mathematics

1 answer:

vekshin12 years ago

6 0

The equation of the line that passes through the given points is y= -4x+23.

<h3>Linear Function</h3>

A line can be represented by a linear function. The standard form for the linear equation is: ax+b , for example, y=2x+7. Where:

a= the slope

b=constant term that represents the y-intercept.

From the stardard form equation:

For point (5,3) : 3=5a+b ( equation 1)
For point (9,-13) : -13=9a+b (equation 2)

Solving the system for equations 1 and 2.

3=5a+b

-13=9a+b * (-1)

3=5a+b

13=-9a-b * (-1)

16= -4a

4= -a

a= -4

If a= -4 from equation 1, you have

3= 5 * (-4)+b

3= -20+b

b=23

Therefore, the line equation is y= -4x+23.

Read more about the system linear equations here:

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Whats is 3/5 as a percent

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3 years ago

Read 2 more answers

2(x+3)^2+2(y)^2=32 what is the center point of the circle and how long is the radius

spayn [35]

Answer:

center (-3, 0)
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Step-by-step explanation:

The standard form equation of a circle is ...

(x -h)^2 +(y -k)^2 = r^2 . . . . . . . . . center at (h, k), radius r

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3 years ago

given the recursive formula for a geometric sequence find the common ratio the 8th term and the explicit formula.did I set these

lesya [120]

Answer:

Step-by-step explanation:

1)Since we know that recursive formula of the geometric sequence is

$a_{n}=a_{n-1}*r$

so comparing it with the given recursive formula $a_{n}=a_{n-1}*-4$

we get common ratio =-4

8th term= $a_{1}*(r)^{n-1}=-2*(-4)^{7} =32768.$

Explicit Formula = $-2*(-4)^{n-1}$

2) Comparing the given recursive formula $a_{n}=a_{n-1}*-2$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-2

8th term= $a_{1}*(r)^{n-1}=-4*(-2)^{7} =512.$

Explicit Formula = $-4*(-2)^{n-1}$

3)Comparing the given recursive formula $a_{n}=a_{n-1}*3$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =3

8th term= $a_{1}*(r)^{n-1}=-1*(3)^{7} =-2187.$

Explicit Formula = $-1*(3)^{n-1}$

4)Comparing the given recursive formula $a_{n}=a_{n-1}*-4$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-4

8th term= $a_{1}*(r)^{n-1}=3*(-4)^{7} =-49152.$

Explicit Formula = $3*(-4)^{n-1}$

5)Comparing the given recursive formula $a_{n}=a_{n-1}*-4$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-4

8th term= $a_{1}*(r)^{n-1}=-4*(-4)^{7} =65536.$

Explicit Formula = $-4*(-4)^{n-1}$

6)Comparing the given recursive formula $a_{n}=a_{n-1}*-2$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-2

8th term= $a_{1}*(r)^{n-1}=3*(-2)^{7} =-384.$

Explicit Formula = $3*(-2)^{n-1}$

7)Comparing the given recursive formula $a_{n}=a_{n-1}*-5$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-5

8th term= $a_{1}*(r)^{n-1}=4*(-5)^{7} =-312500.$

Explicit Formula = $4*(-5)^{n-1}$

8)Comparing the given recursive formula $a_{n}=a_{n-1}*-5$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-5

8th term= $a_{1}*(r)^{n-1}=2*(-5)^{7} =-156250.$

Explicit Formula = $2*(-5)^{n-1}$

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3 years ago

Solve the equation : 2x + 7 = 21 and find the value of x. Is 2x + 7 = 21 a first degree equation or not?

Paraphin [41]

Answer:

The proposed equation is solved as follows:

2X + 7 = 21

2X = 21 - 7

2X = 14

X = 14 / 2

X = 7

The value of X is 7.

A first degree equation is an algebraic equation in which each term is either a constant or a product of a fixed term on a single variable. Therefore, this equation is a first degree one, since it only has a single variable, which is X.

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2 years ago