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Montano1993 [528]
3 years ago
14

Write an equation of the line that passes through a pair of points:

Mathematics
1 answer:
vredina [299]3 years ago
3 0

Answer:

b)

The  equation of the line that passes through a pair of points

   y  = \frac{-1}{8}  ( x ) -\frac{11}{8}

Step-by-step explanation:

 Explanation:-

Given points are ( 5 ,-2) and ( 3,-1)

slope of the line

                        m = \frac{y_{2} -y_{1} }{x_{2}-x_{1}  } = \frac{-1-(-2)}{-3-(5)} = \frac{1}{-8}

The  equation of the line that passes through a pair of points

              y - y_{1} = m ( x - x_{1} )

           y - (-2) = \frac{-1}{8}  ( x - (5) )

          y  = \frac{-1}{8}  ( x ) +\frac{5}{8} -2

        y  = \frac{-1}{8}  ( x ) +\frac{5-16}{8}

      y  = \frac{-1}{8}  ( x ) -\frac{11}{8}

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Which is the graph of g(x) = [X + 3]
Black_prince [1.1K]

Answer:

The "top center" one

Step-by-step explanation:

-y=x+3

-Use the <em>slope-intercept form</em> to find the slope and y-intercept. (y=mx+b)

m=1 b=3

-Slope: 1 , y-intercept: 3

X   | y

-3 | 0

 0 | 3

Then you can use a graphing calculator if you'd like !

6 0
3 years ago
How many lbs of peanuts must mike add to 9 lbs of mixed nuts containing 40% peanuts to make a mixture with 73% peanuts
andreev551 [17]

The new mixture exists 56.5% peanuts.

<h3>How to estimate the total number of peanuts required for the mixture?</h3>

The first batch of 9 lb of mixed nuts possesses 40% peanuts.

This means the quantity of peanuts exists:

9 * 40/100 = 3.6 lb

The second batch of 9 lb of mixed nuts possesses 73% peanuts.

This means the quantity of peanuts exists:

9 * 73/100 = 6.57 lb

The total quantity of peanuts in the mix exists

3.6 lb + 6.57 lb = 10.17 lb

There exist 9 lb + 9 lb = 18 lb of mix.

Therefore, the percent of peanuts in the mix exists:

10.17 / 18 * 100 = 56.5%

The new mixture exists 56.5% peanuts.

To learn more about the total quantity refer to:

brainly.com/question/18919760

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3 0
2 years ago
Luis has 3 1/3 yards of ribbon to use for two separate presents, one of which is very large. If Luis uses 2 1/2 yards to decorat
never [62]

Answer:

\frac{5}{6} yards

Step-by-step explanation:

Luis had 3\frac{1}{3} yards of ribbon in total to use for two presents. For the first present Luis used 2\frac{1}{2} yards of the ribbon.

Since she had 3\frac{1}{3} yards in total and she used  2\frac{1}{2} yards from it, the amount of ribbon left for second present can be obtained by taking out(subtracting)  2\frac{1}{2} yards from 3\frac{1}{3}.

So, the number of yards of ribbon left for second present will be:

3\frac{1}{3}-2\frac{1}{2}\\\\ =\frac{10}{3}-\frac{5}{2}\\\\ =\frac{10(2)-5(3)}{2 \times 3}\\\\ =\frac{20-15}{6}\\\\ =\frac{5}{6}

This means Luis had \frac{5}{6} yards of ribbon for the second present.

5 0
4 years ago
Help thankss <br>answer is <br><img src="https://tex.z-dn.net/?f=%20%28%20-%2014.0%29" id="TexFormula1" title=" ( - 14.0)" alt="
lyudmila [28]
We have point P=(x_0,y_0)=(2,-4), so first calculate f'(x_0). There will be:

y=f(x)=2x^3-5x^2\\\\\\\\f'(x)=\big(2x^3-5x^2\big)'=\big(2x^3\big)'-\big(5x^2\big)'=2\big(x^3\big)'-5\big(x^2\big)'=\\\\=2\cdot3x^2-5\cdot2x=\boxed{6x^2-10x}\\\\\\\\f'(x_0)=f'(2)=6\cdot2^2-10\cdot2=6\cdot4-20=24-20=\boxed{4}

Now, we can write the equation of the normal line as:

y-y_0=-\dfrac{1}{f'(x_0)}(x-x_0)\\\\\\ y-(-4)=-\dfrac{1}{4}(x-2)\\\\\\y+4=-\dfrac{1}{4}x+\dfrac{1}{2}\\\\\\y=-\dfrac{1}{4}x+\dfrac{1}{2}-4\\\\\\\boxed{y=-\dfrac{1}{4}x-\dfrac{7}{2}}

Normal line (and every line) crosses x-axis when y = 0, so coordinates of A:

\boxed{y=0}\qquad\qquad\text{and}\\\\\\y=-\dfrac{1}{4}x-\dfrac{7}{2}\\\\\\&#10;0=-\dfrac{1}{4}x-\dfrac{7}{2}\\\\\\\dfrac{1}{4}x=-\dfrac{7}{2}\quad|\cdot4\\\\\\x=-\dfrac{7\cdot4}{2}\\\\\\x=-7\cdot2\\\\\\\boxed{x=-14}\\\\\\\boxed{A=(-14,0)}
8 0
3 years ago
There are 1,000,000 microseconds in a second. If the length of a day on a planet increases by 1,250 microseconds
densk [106]

Answer:

The day would be 3.125 seconds longer in 25 centuries.

Step-by-step explanation:

Let's break it down.

Each year, 1250 microseconds are added to the planet's day.

We need to find how long a day would be in 25 centuries.

A century is 100 years.

A second is 1,000,000 microseconds.

100*25=2500\\\\2500*1250=3125000\\\\\frac{3125000}{1000000}=\frac{3125}{1000}=\frac{3.125}{1}=3.125

The day would be 3.125 seconds longer in 25 centuries.

3 0
3 years ago
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