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2 years ago

Find an equation for a line that passes through points -4,3 and 6,5

2 answers:

7 0

First you would want to find the slope. The equation for slope is y2-y1/x2-x1.

So the slope would be 5-3/ 6-(-4). Two negatives make a positive so it would be 5-3/6+4 or 2/10 which can be simplified to 1/2.

Next you used the point slope form equation. y-y1 =m(x-x1)

m is the slope we just found: 1/2
the equation would be = y-3=1/2(x-(-4))

two negatives make a positive so the equation would now be y-3=1/2(x+4)

You then distribute the x to what is in the parenthesis

y-3=1/2x+2
then add the 3 to both sides

you’re equation should be y=1/2+5

pickupchik [31]2 years ago

3 0

Answer: Find the equation of the line that passes through the points (-6,2) and (-5,3). Write your answer in standrard form. Answer by fractalier(6550) ...

Step-by-step explanation:

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Fraction strip for 1/4, 2/8,3/12

Brilliant_brown [7]

4/16, 5/20, 6/24, 7/28, 8/32...

7 0

2 years ago

there were twenty one adults in line at a movie theater.That is three times the number of children in line. how many children we

V125BC [204]

You would do 21 divided by 3 and get 7.

The answer is 7 kids are in the line.

Please vote my answer the brainiest if it was helpful!!

7 0

3 years ago

For the function y=3x2: (a) Find the average rate of change of y with respect to x over the interval [3,6]. (b) Find the instant

nirvana33 [79]

Answer:

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

Step-by-step explanation:

a) Geometrically speaking, the average rate of change of $y$ with respect to $x$ over the interval by definition of secant line:

$r = \frac{y(b) -y(a)}{b-a}$ (1)

Where:

$a$ , $b$ - Lower and upper bounds of the interval.

$y(a)$ , $y(b)$ - Function exaluated at lower and upper bounds of the interval.

If we know that $y = 3\cdot x^{2}$ , $a = 3$ and $b = 6$ , then the average rate of change of $y$ with respect to $x$ over the interval is:

$r = \frac{3\cdot (6)^{2}-3\cdot (3)^{2}}{6-3}$

$r = 27$

The average rate of change of $y$ with respect to $x$ over the interval $[3,6]$ is 27.

b) The instantaneous rate of change can be determined by the following definition:

$y' = \lim_{h \to 0}\frac{y(x+h)-y(x)}{h}$ (2)

Where:

$h$ - Change rate.

$y(x)$ , $y(x+h)$ - Function evaluated at $x$ and $x+h$ .

If we know that $x = 3$ and $y = 3\cdot x^{2}$ , then the instantaneous rate of change of $y$ with respect to $x$ is:

$y' = \lim_{h \to 0} \frac{3\cdot (x+h)^{2}-3\cdot x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{(x+h)^{2}-x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{2\cdot h\cdot x +h^{2}}{h}$

$y' = 6\cdot \lim_{h \to 0} x +3\cdot \lim_{h \to 0} h$

$y' = 6\cdot x$

$y' = 6\cdot (3)$

$y' = 18$

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

5 0

2 years ago

Find the measure of the smallest angle?

weeeeeb [17]

Answer:

m ∠RMK = 51°

Step-by-step explanation:

m ∠JMK = m ∠RMK + m ∠JMR

10x + 19 = 7x - 26 + 6x + 12

10x +19 = 13x -14

19 = 3x -14

33 = 3x

11 = x

m ∠RMK = 7(11) - 26 = 51°

m ∠JMR = 6 (11) + 12 = 78

6 0

2 years ago

How do I graph g(x)=f(x)+5

sashaice [31]

Find the absolute value vertex. In this case, the vertex for y=|x−5|y=|x-5| is (5,0)(5,0).

(5,0)(5,0)

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

(−∞,∞)(-∞,∞)

{x|x∈R}{x|x∈ℝ}

For each xx value, there is one yy value. Select few xx values from the domain. It would be more useful to select the values so that they are around the xx value of the absolute valuevertex.

xy3241506172

8 0

2 years ago