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dalvyx [7]
3 years ago
13

A line passes through the points (2,4) and (5,6) .

Mathematics
1 answer:
Alenkinab [10]3 years ago
8 0

Answer:

Option B and D are correct.

Step-by-step explanation:

Given: A line passes through the points (2,4) and (5,6).

* Case 1:

If a line passes through the points (2, 4) and (5, 6)

Point slope intercept form:

for any two points (x_1,y_1) and (x_2, y_2)

then the general form y -y_1=m(x-x_1) for linear equations where m is the slope given by:

m =\frac{y_2-y_1}{x_2-x_1}

First calculate slope for the points (2, 4) and (5, 6);

m = \frac{y_2-y_1}{x_2-x_1} =\frac{6-4}{5-2} = \frac{2}{3}

then, by point slope intercept form;

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

* Case 2:

If a line passes through the points (5, 6) and (2, 4)

First calculate slope for the points (5, 6) and (2, 4);

m = \frac{y_2-y_1}{x_2-x_1} =\frac{4-6}{2-5} = \frac{-2}{-3}= \frac{2}{3}

then, by point slope intercept form;

y-6=\frac{2}{3}(x-5)

Yes, the only equation of line from the given options  which describes the given line are;

y-4=\frac{2}{3}(x-2)  and y-6=\frac{2}{3}(x-5)



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Answer:

C

Step-by-step explanation:

We are given

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We can see that first term is 1.25x

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3 years ago
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. Use the quadratic formula to solve each quadratic real equation. Round
Liono4ka [1.6K]

Answer:

A. No real solution

B. 5 and -1.5

C. 5.5

Step-by-step explanation:

The quadratic formula is:

\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}, with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}

\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5

So B has two solutions of 5 and -1.5.

Now to C!

\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}

\frac{44}{8} = 5.5

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

4 0
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8 0
3 years ago
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Ganezh [65]
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now that you know b
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so the answer is 
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5 0
2 years ago
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