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3 years ago

12

9n+21−3n−24 simplify

1 answer:

Thepotemich [5.8K]3 years ago

6 0

Alright First you combine the like terms.
9n - 3n = 6n
21 - 24 = -3

Rewrite the simplified equation
6n - 3 = 0

Solve for n
6n = 3
n = 3/6
n = 1/2

Ask if you need more help! :)

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Not 6 not 5 not 8.46

kozerog [31]

Answer:

The answer is square root 72 or if you don't want to put it under the square root, just simplify and you get 8.49. Just use the distance formula.

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14. What's the length of a segment that begins at the point (2, 3) and ends at the point (18, 15)?

Bingel [31]

Answer:

20

Step-by-step explanation:

d = √(xb - xa)2 + (yb - ya)2 =

= √(18 - 2)2 + (15 - 3)2 =

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3 years ago

15.36 x (-0.45) helpp

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3 years ago

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the pa

konstantin123 [22]

Answer:

$y=2x-8$

Step-by-step explanation:

The given parametric equation is;

$x=6+ln(t),y=t^2+3$

<h3><u>BY ELIMINATING THE PARAMETER</u></h3>

To eliminate the parameter we make $t$ the subject in one equation and put it inside the other.

We make $t$ the subject in $x=6+ln(t)$ because it is easier.

$\Rightarrow x-6=ln(t)$

$\Rightarrow {e}^{x-6}=e^{ln(t)}$

$\Rightarrow {e}^{x-6}=t$

Or

$t={e}^{x-6}$

We now substitute this into $y=t^2+3$ .

This gives us;

$y=(e^{x-6})^2+3$ .

$\Rightarrow y=e^{2(x-6)}+3$ .

We have now eliminated the parameter.

The equation of the tangent at (6,4) is given by;

$y-y_1=m(x-x_1)$

where the gradient function is given by;

$\frac{dy}{dx}=2e^{2(x-6)}$

We substitute $x=6$ into the gradient function to obtain the gradient.

$\Rightarrow m=2e^{2(6-6)}$

$\Rightarrow m=2e^0$

$\Rightarrow m=2$

The equation of the tangent becomes

$y-4=2(x-6)$

We simplify to obtain

$y=2x-12+4$

$y=2x-8$

<h3><u>WITHOUT ELIMINATING THE PARAMETER</u></h3>

The given parametric equation is;

$x=6+ln(t),y=t^2+3$

For $x=6+ln(t)$

$\frac{dx}{dt}=\frac{1}{t}$

For $y=t^2+3$

$\frac{dy}{dt}=2t$

The slope is given by;

$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\frac{dy}{dx}=\frac{2t }{\frac{1}{t} }$

$\frac{dy}{dx}=2t^2$

At the point, (6,4), we plug in any of the values into the parametric equation and find the corresponding value for $t$ .

Notice that

When $x=6$ , $6=6+\ln(t)$

$6-6=\ln(t)$

$0=\ln(t)$

$e^0=e^\ln(t)$

$1=t$

when $y=4$ , $4=t^2+3$

$4-3=t^2$

$1=t^2$

$t=\pm1$

But the slope is the same when we plug in any of these values for t.

$\frac{dy}{dx}=2(\pm1)^2=2$

The equation of the tangent becomes

$y-4=2(x-6)$

We simplify to obtain

$y=2x-12+4$

$y=2x-8$

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3 years ago

The diameter of Saturn at its equator is about 120,540,000 meters. What is 120,540,000 written in word form?

Agata [3.3K]

Answer:

1) one hundred twenty million five hundred forty thousand

2) 2,040 (equation: 4(5 x 102))

3) 1,872

4) 9.25, 9.325, 9.5, 9.675

5) 18.4

6) 108 (equation: 3(21) + 3(15))

7) $0.95

8) $2.50

Step-by-step explanation:

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3 years ago