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Vilka [71]
4 years ago
6

Graph ​ y−2=−34(x−6) ​ using the point and slope given in the equation.

Mathematics
2 answers:
prohojiy [21]4 years ago
8 0
The points are at (8,0) and (0,6)
Novay_Z [31]4 years ago
8 0

Answer:

Point-slope form: An equation of a straight line in the form y -y_1 = m(x -x_1);  

where

m is the slope of the line and (x_1, y_1) are the coordinates of a given point on the line.

Given the equation: y-2=-\frac{3}{4}(x-6)         ......[1]

On comparing with Point slope form equation we have;

m = -\frac{3}{4} and point (6 , 2)

Now, find the Intercept of the given equation:

x-intercept: The graph crosses the the x-axis i.e,

Substitute y =0 in [1] and solve for x;

0-2=-\frac{3}{4}(x-6)

-2=-\frac{3}{4}(x-6)

Using distributive property:  a\cdot(b+c) = a\cdot b +a\cdot c

-2 = -\frac{3}{4}x + \frac{18}{4}

Subtract \frac{18}{4} on both sides we get;

-2-\frac{18}{4}= -\frac{3}{4}x + \frac{18}{4} -\frac{18}{4}

Simplify:

-\frac{26}{4} = -\frac{3}{4}x

or

-26 = -3x

Divide both sides by -3 we get;

x = 8.667

x-intercept: (8.667, 0)

Similarly, for

y-intercept:

Substitute x = 0 in [1] and solve for y;

y-2=-\frac{3}{4}(0-6)

y-2=\frac{18}{4}

Add 2 on both sides we get;

y-2+2=\frac{18}{4}+2

Simplify:

 y=\frac{26}{4} =6.5

y-intercept: (0, 6.5)

Now, using these two points (8.667, 0) and (0, 6.5) you can plot the graph using line tool as shown below.


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Answer:

Each piece of tofu weighs 0.847 ounces.

Step-by-step explanation:

8.47 divided by 10 = 0.847 ounces

Hope this helps

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41

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Write an equation of the line that passes through the given point and is parallel to the graph of the given equation.
True [87]
Question 1:

--------------------------------------------------------------------
Find Slope
--------------------------------------------------------------------
Equation: y = 5x - 2
Slope = 5
Slope of parallel line = 5

--------------------------------------------------------------------
Insert slope into the general equation y = mx + c
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y = 5x + c

--------------------------------------------------------------------
Find y-intercept
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At point (2, -1)
y = 5x + c
-1 = 5(2) + c
c = -1 - 10
c = -11

--------------------------------------------------------------------
Insert y-intercept into the equation
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y = 5x + c
y = 5x - 11

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Answer: y = 5x - 11
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Question 2:

--------------------------------------------------------------------
Find Slope
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y = 9x 
Slope = 9 
Slope of the parallel line = 9

--------------------------------------------------------------------
Insert slope into the equation y = mx + c
--------------------------------------------------------------------
y = 9x + c

--------------------------------------------------------------------
Find y-intercept
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y = 9x + c
At point (0, 5)
5 = 9(0) + c
c = 5

--------------------------------------------------------------------
Insert y-intercept into the equation
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y = 9x + c
y = 9x + 5

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Answer: y = 9x + 5
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7 0
3 years ago
Find the domain of the function y = 3 tan(23x)
solmaris [256]

Answer:

\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

In other words, the x in f(x) = 3\, \tan(23\, x) could be any real number as long as x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) for all integer k (including negative integers.)

Step-by-step explanation:

The tangent function y = \tan(x) has a real value for real inputs x as long as the input x \ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

Hence, the domain of the original tangent function is \mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

On the other hand, in the function f(x) = 3\, \tan(23\, x), the input to the tangent function is replaced with (23\, x).

The transformed tangent function \tan(23\, x) would have a real value as long as its input (23\, x) ensures that 23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

In other words, \tan(23\, x) would have a real value as long as x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right).

Accordingly, the domain of f(x) = 3\, \tan(23\, x) would be \mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

4 0
3 years ago
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