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Vladimir79 [104]
3 years ago
12

Find the equation of a line that passes through the points (6,7) and (8,3).

Mathematics
1 answer:
horsena [70]3 years ago
3 0

Answer:

y=-2x+19

Step-by-step explanation:

For lines like these, the slope is always defined as "the change in y over the change in x" or, in equation form:  m= \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }.

Now we have,

m= 3-7/8-6

m=-2. Since m=-2, the equation is y=-2x+b.

To find b, think about what your (x,y) points mean:

(6,7). When x of the line is 6, y of the line must be 7.

(8,3). When x of the line is 8, y of the line must be 3.

Substitute them to the equation  y=-2x+b.

You can use either (x,y) point you want..the answer will be the same:

(6,7). y=mx+b or 7=-2 × 6+b, or solving for b: b=7-(-2)(6). b=19.

(8,3). y=mx+b or 3=-2 × 8+b, or solving for b: b=3-(-2)(8). b=19.

So, the equation of the line that passes through the points

(6,7) and (8,3)  is  y=-2x+19.

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What is the answer and how do you solve <br><br><br> 21x^3/3x^−1
Sunny_sXe [5.5K]

Answer:

  as written: 7x^2

  perhaps intended: 7x^4

Step-by-step explanation:

As written, the Order of Operations specifies that the only denominator factor is 3, so the expression is interpreted as ...

  \dfrac{21x^3}{3}x^{-1}=\dfrac{21}{3}x^{3-1}=7x^2

___

Sometimes grouping of denominator factors is intended, but parentheses are not written as they should be. If the intention is ...

  21x^3/(3x^-1)

then the simplification is ...

  \dfrac{21x^3}{3x^{-1}}=\dfrac{21}{3}x^{3-(-1)}=7x^4

_____

The applicable rules of exponents are ...

  (a^b)(a^c) = a^(b+c)

  a^b/a^c = a^(b-c)

6 0
3 years ago
Describe and correct the error a student made when graphing y+5=-3/4(x-8).
Elis [28]
In step 1 the y-intercept should be plotted at (0,-6)
Step-by-step explanation:
Remember that to find the Y intercept in any linear equation you need to use 0 as your X value, this means taking the formula in the y=mx+b form and replacing X with a 0.
Since the formula is y = -3/4x -6
We just insert a 0 insted of the "x"
y = -3/4(0) -6
y=0-6
y=6
So the y-intercept sould be placed in (0,-6)
That's what he did wrong when graphing the equation.
8 0
2 years ago
Solve the quadratic equation by completing the square. x^2+14x+47=0. What is the form and solution?
Phoenix [80]
To complete a square for ax²+bx+c, you first make a=1.
in this case, a is already 1
move c to the right side: x²+14x=-47
next, add (1/2 of b)² to both sides, in this case, b=14, half of 14 is 7, so x²+14x+ 7²=-47+7²
the left side is now a perfect square: (x+7)²=2
x+7=√2 or x+7=-√2
x=√2-7, or x=-2-√7
4 0
3 years ago
Helppppppppppppppppp
barxatty [35]

Answer:

This looks hard.

Step-by-step explanation:

I know because it is very hard.

4 0
2 years ago
The term "freshman 15" refers to the claim that college students typically gain 15lbs during freshman year at college. Assume th
Viefleur [7K]

The probability that a randomly selected male college student gains 15 lb or more during their freshman year is 11.6%

<h3>What is Probability ?</h3>

Probability is defined as the likeliness of an event to happen.

Let X be a random variable that shows the term "freshman 15" that claims that students typically gain 15lb during their freshman year at college.

It is given that

X follows is a normal distribution with a mean of 2.1 lb (μ) and a standard deviation (σ)  of 10.8 lb.

Population Mean (μ) = 2.1

Population Standard Deviation (σ) = 10.8

We need to compute Pr(X≥15). The corresponding z-value needed to be computed is:

\rm Z_{lower} = \dfrac{ X_1 -\mu }{\sigma}\\\\Z_{lower} = \dfrac{ 15-2.1 }{10.8}\\\\\\Z_{lower} = 1.19

Then the probability is given as

\rm Pr(X \geq 16 ) = Pr (\dfrac{X -21}{10.8} \geq \dfrac{15-21}{10.8})\\\\= Pr (Z \geq \dfrac{15-2.1}{10.8}\\\\= Pr (Z\geq 1.19)\\\\ = 0.1162

Pr(X≥15)=0.1162. (11.6%)

The probability that a randomly selected male college student gains 15 lb or more during their freshman year is 11.6%

To know more about Probability

brainly.com/question/11234923

#SPJ1

3 0
2 years ago
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