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3 years ago

The table represents a linear function. Which is the slope of the function?

Mathematics

2 answers:

Galina-37 [17]3 years ago

7 0

From the table we have the points (-4, -2) and (-2, -10).

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

Substitute:

$m=\dfrac{-10-(-2)}{-2-(-4)}=\dfrac{-10+2}{-2+4}=\dfrac{-8}{2}=-4$

<h3>Answer: The slope = -4.</h3>

seropon [69]3 years ago

5 0

Answer:

I think that the answer is -4

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Has endpoints J(13, 8) and K(–6, –6). Find the coordinates of the midpoint of .

vova2212 [387]

Answer:

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Step-by-step explanation:

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2 years ago

Find the 12th term of the geometric sequence 1, 2,4

irakobra [83]

Answer:

The 12th term in the sequence is 22

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3 years ago

Read 2 more answers

The equation $a^7xy-a^6y-a^5x=a^4(b^4-1)$ is equivalent to the equation $(a^mx-a^n)(a^py-a^2)=a^4b^4$ for some integers $m$, $n$

Ksivusya [100]

Let <span>simplify the equations $a^7xy-a^6y-a^5x=a^4(b^4-1)$ and $(a^mx-a^n)(a^py-a^2)=a^4b^4$ :
</span>
<span>1) $a^7xy-a^6y-a^5x+a^4=a^4b^4$
</span>
<span>and
</span>
<span>2) $a^{m+p}xy-a^{n+p}y-a^{m+2}x+a^{n+2}=a^4b^4$ .
</span>
<span>Equate the coefficients:
</span><span>
</span><span> $xy: m+p=7 \\ y: n+p=6 \\ x: m+2=5 \\ 1: n+2=4$ .</span>
Then $n=2 \\ p=4 \\ m=3$ and mnp=24.
<span />

3 0

3 years ago

Select the THREE equations that this diagram could represent.

Vladimir [108]

<u>Question </u>

Select the three equations that this diagram could represent.

<u>Answer</u>

<em>Well, we first have to find out what the diagram says.</em>

<em>So what the diagram says is that 18 + 18 + 18 = 54.</em>

<em>Given this information, we have to figure what other answers = 54.</em>

<em>Therefore the answers are </em>

(A) 18 * 3 = 54.

(D) 54/18 = 3

(E) 54/3 =18

6 0

3 years ago

Find the radius of the circle with equation x²+y²+8x+8y+28=0

galina1969 [7]

<h2>Hello!</h2>

The answer is:

Center: (-4,-4)

Radius: 2 units.

To solve the problem, using the given formula of a circle, we need to find its standard equation form which is equal to:

$(x-h)^{2}+(y-k)^{2}=r^{2}$

Where:

"h" and "k" are the coordinates of the center of the circle and "r" is its radius.

So, we need to complete the square for both variable "x" and "y".

The given equation is:

$x^{2}+y^{2}+8x+8y+28=0$

So, solving we have:

$x^{2}+y^{2}+8x+8y=-28$

$(x^2+8x+(\frac{8}{2})^{2})+(y^2+8y+(\frac{8}{2})^{2})=-28+((\frac{8}{2})^{2})++(\frac{8}{2})^{2})\\\\(x^2+8x+16 )+(y^2+8y+16)=-28+16+16\\\\(x^2+4)+(y^2+4)=4$

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

Now, we have that:

$h=-4\\k=-4\\r=\sqrt{4}=2$

So,

Center: (-4,-4)

Radius: 2 units.

Have a nice day!

Note: I have attached a picture for better understanding.

3 0

3 years ago