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3 years ago

Which is an equation in point-slope form for the given point and slope?

Mathematics

1 answer:

xxMikexx [17]3 years ago

4 0

Hi There!

Solution,

Point Slope Form: y - y1 = m(x - x1)

Now you have a reference equation of what the answer should look like.

By looking at the answer choices you can eliminate A.

Tip: When you have a negative number you are going to add. For Example, if you have -8, you will input x + 8 or y + 8.

Tip 2: When you have a positive number you are going to subtract. For Example, if you have 3 you will input x - 3 or y - 3.

We can eliminate B because its not subtracting 3.

We can also eliminate C because its not adding -8.

The best answer choice is D.

Answer,

y – 3 = 6(x + 8)

I hope you understand, so you can do this in the future on your own.

Hope This Helps :)

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wel

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Step-by-step explanation:

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3 years ago

Pliz solve this question

kotegsom [21]

Answer:

i) (-3,0)
ii) (0,5)

Step-by-step explanation:

<u>Coordinates of points</u>

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It lies on y-axis, it has zero x-coordinate
It is (0, 5)

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3 years ago

What is the missing term in the quadratic expression below (2x-3)(x+4)=2x^2+ ___ - 12

Allushta [10]

5x is the missing term

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3 years ago

What is x multipied by -x

Alik [6]

The result of x*(-x)= $-x^2$

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3 years ago

Read 2 more answers

If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t

sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, $6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296$

This implies that $6^{4}$ exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, $6^{3}=216$ and $216\times 5=1080, 216\times 4=864$ . This implies that $216\times 5$ exceeds 1052 and thus there can be at most 4 groups of $6^{3}$ .

Then,

$1052-4\times6^{3}$

$1052-4\times216$

$1052-864$

$188$

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, $6^{2}=36$ and $36\times 5=180, 36\times 6=216$ . This implies that $36\times 6$ exceeds 188 and thus there can be at most 5 groups of $6^{2}$ .

Then,

$188-5\times6^{2}$

$188-5\times36$

$188-180$

$8$

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, $6^{1}=6$ and $6\times 1=6, 6\times 2=12$ . This implies that $6\times 2$ exceeds 8 and thus there can be at most 1 group of $6^{1}$ .

Then,

$8-1\times6^{1}$

$8-1\times6$

$8-6$

$2$

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, $6^{0}=1$ and $1\times 2=2$ . This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

Then,

$1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}$

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

6 0

3 years ago