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3 years ago

Consider the following equation. Graph 5x + 6y = 30 PLEASE SHOW STEPS

Mathematics

1 answer:

MissTica3 years ago

6 0

Answer:

Step-by-step explanation:

Convert the equation into slope-intercept form.

5x-6y =30 Subtract 5x from each side

-6y = -5x + 30 Multiply each term by -1

6y = 5x – 30 Divide each side by 6

y = ⅚x - 5

2. Create a table of different x and y values,

x ⅚x – 5 y

-12 ⅚(-12) – 5 -15

-6 ⅚(-6) – 5 -10

0 ⅚(0) – 5 -5

6 ⅚(6) – 5 0

12 ⅚(12) – 5 5

3. Convert the table to ordered pairs

(-12,-15)

(-6, -10)

(0, -5)

(6, 0)

(12, -5)

4. Plot the points and draw a line through them.

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Two points are _______________________ collinear

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Step-by-step explanation:

Two points are always collinear since we can draw a distinct (one) line through them.

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3 years ago

Suppose that $p$ and $q$ are positive numbers for which \[\log_9 p = \log_{12} q = \log_{16} (p + q).\] what is the value of $q/

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Given $p,q>0$ and $\log_9p=\log_{12}q=\log_{16}(p+q)=x$ (say).

Then,

$p=9^x\\ q=12^x\\ p+q=16^x$

From the above 3 equations,

$\frac{q}{p} =(\frac{12}{9} )^x\\ \frac{q}{p} =(\frac{4}{3} )^x\\ \frac{p+q}{p} =(\frac{16}{9} )^x\\ \frac{p+q}{p} =(\frac{4}{3} )^{2x}\\$

From the equations, we get

$\frac{p+q}{p}=(\frac{q}{p})^2\\ 1+\frac{q}{p}=(\frac{q}{p})^2\\ (\frac{q}{p})^2-\frac{q}{p}-1=0\\ \frac{q}{p}=\frac{1 \pm \sqrt{5}}{2}$

Since $p,q>0$ , the negative value is rejected.

$\frac{q}{p}=\frac{1 + \sqrt{5}}{2}$

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