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3 years ago

S=a+(n-1)d,solve for d

Mathematics

1 answer:

Vinvika [58]3 years ago

5 0

Answer:\frac{s-a}{n-1}

Step-by-step explanation:

formula, s=a+(n-1)d

1. get rid of the a on d's side by subtracting it from both sides.

s=a+(n-1)d becomes s-a=(n-1)d

2. get rid of (n-1) by dividing it from both sides. s-a=(n-1)d becomes

\frac{s-a}{n-1} =d

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Step-by-step explanation:

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4 years ago

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Simora [160]

Answer:

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Step-by-step explanation:

This question is based on Secant Secant theorem.

Secant Secant theorem gives us the following formula:

(AB + BD)AB = (AC + CE).AC

From the above question we have the following parameters

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3 years ago

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nataly862011 [7]

Answer:

option d

{ 4, 8 , 16 , 32 }

Step-by-step explanation:

Given in the question a function

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$x_{1}=4$

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3 years ago

A planet follows an elliptical path described by 16 x squared + 4 y squared = 64. A comet follows the parabolic path y equals x

saul85 [17]

Answer:

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Step-by-step explanation:

Given that, a planet follows an elliptical path described by

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A comet follows the parabolic path

$y=x^2-4$

$x^2=y+4$ .........(2)

To find the intersecting point, we need to solve the above equations.

Putting $x^2=y+4$ in the equation (1)

$y+4+4y^2=64$

$\Rightarrow 4y^2+y+4-64=0$

$\Rightarrow 4y^2+y-60=0$

$\Rightarrow 4y^2+16y-15y-60=0$

$\Rightarrow 4y(y+4)-15(y+4)=0$

$\Rightarrow(y+4)(4y-15)=0$

$\Rightarrow y=-4, \frac{15}{4}$

When y = -4 , then $x^2=-4+4$ $\Rightarrow x=0$

When $y=\frac{15}{4}$ , then $x^2=\frac{15}{4}+4$ $\Rightarrow x=\pm \frac{\sqrt{31}}{2}$

The comet might intersect the orbiting planet at (0, -4), $(\pm\frac{\sqrt{31}}{2},\frac{15}{4})$ .

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3 years ago

S=a+(n-1)d,solve for d​

S=a+(n-1)d,solve for d