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3 years ago

Find the slope of the line that passes through the given

Mathematics

1 answer:

Anton [14]3 years ago

6 0

Answer:

Step-by-step explanation:

4x − 3= −12x + 13

4x + 12x = 13 + 3

16x = 16

x = 1

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PLEASE HELP!

emmainna [20.7K]

Use the given recursion and starting value of $x_0 = 2.4$ to find $x_1$ :

$x_1 = \dfrac{x_0 + \frac6{x_0}}2 = \dfrac{2.4 + \frac{6}{2.4}}2 = 2.45$

Do the same for $x_2$ and $x_3$ :

$x_2 = \dfrac{x_1 + \frac6{x_1}}2 = \dfrac{2.45 + \frac6{2.45}}2 \approx 2.44949$

$x_3 = \dfrac{x_2+\frac6{x_2}}2 \approx \dfrac{2.44949 + \frac6{2.44949}}2 \approx \boxed{2.44949}$

(That's not a mistake. This just tells you that the 2nd and 3rd iterates are very close together and have at least the same first 5 digits after the decimal.)

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2 years ago

What is the area of this face?

BARSIC [14]

Answer:

10*4-6*2=28

Step-by-step explanation:

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2 years ago

Read 2 more answers

How many hours and minutes are in 700 minutes

Nadusha1986 [10]

I hope this helped you question.... please mark as brainliest

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3 years ago

What is the answer for 3r+8=2r+12

Marina86 [1]

Answer:

r=4

Step-by-step explanation:

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3 years ago

Read 2 more answers

PLEASE HELP ASAP! I don’t recall how to do this!

MakcuM [25]

Answer:

Step-by-step explanation:

For a. we start by dividing both sides by 200:

$(1.05)^x=1.885$

In order to solve for x, we have to get it out from its position of an exponent. Do that by taking the natural log of both sides:

$ln(1.05)^x=ln(1.885)$

Applying the power rule for logs lets us now bring down the x in front of the ln:

x * ln(1.05) = ln(1.885)

Now we can divide both sides by ln(1.05) to solve for x:

$x=\frac{ln(1.885)}{ln(1.05)}$

Do this on your calculator to find that

x = 12.99294297

For b. we will first apply the rule for "undoing" the addition of logs by multipllying:

$ln(x*x^2)=5$

Simplifying gives you

$ln(x^3)=5$

Applying the power rule allows us to bring down the 3 in front of the ln:

3 * ln(x) = 5

Now we can divide both sides by 3 to get

$ln(x)=\frac{5}{3}$

Take the inverse ln by raising each side to e:

$e^{ln(x)}=e^{\frac{5}{3}}$

The "e" and the ln on the left undo each other, leaving you with just x; and raising e to the power or 5/3 gives you that

x = 5.29449005

For c. begin by dividing both sides by 20 to get:

$\frac{1}{2}=e^{.1x}$

"Undo" that e by taking the ln of both sides:

$ln(.5)=ln(e^{.1x})$

When the ln and the e undo each other on the right you're left with just .1x; on the left we have, from our calculators:

-.6931471806 = .1x

x = -6.931471806

Question d. is a bit more complicated than the others. Begin by turning the base of 4 into a base of 2 so they are "like" in a sense:

$(2^2)^x-6(2)^x=-8$

Now we will bring over the -8 by adding:

$(2^2)^x-6(2)^x+8=0$

We can turn this into a quadratic of sorts and factor it, but we have to use a u substitution. Let's let $u=2^x$

When we do that, we can rewrite the polynomial as

$u^2-6u+8=0$

This factors very nicely into u = 4 and u = 2

But don't forget the substitution that we made earlier to make this easy to factor. Now we have to put it back in:

$2^x=4,2^x=2$

For the first solution, we will change the base of 4 into a 2 again like we did in the beginning:

$2^2=2^x$

Now that the bases are the same, we can say that

x = 2

For the second solution, we will raise the 2 on the right to a power of 1 to get:

$2^x=2^1$

Now that the bases are the same, we can say that

x = 1

5 0

3 years ago