Give 3 solutions on 14x+7y=28 that will satisfy the consistent systems

Mathematics

2 answers:

Ronch [10]3 years ago

5 0

X=1, y=2
x=0, y= 4
x=2, y=0

Sav [38]3 years ago

4 0

Answer:

Three solutions =

x=2, y=0

x= 0, y=4

x=1, y=2

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Hey if you see this i really need your help its an exam thanks

professor190 [17]

Answer:

I think it is Letter B

Because I add the two

L-2, +N 12=10

7 0

2 years ago

3^x= 3*2^x solve this equation

kompoz [17]

In the equation

$3^x = 3\cdot 2^x$

divide both sides by $2^x$ to get

$\dfrac{3^x}{2^x} = 3 \cdot \dfrac{2^x}{2^x} \\\\ \implies \left(\dfrac32\right)^x = 3$

Take the base-3/2 logarithm of both sides:

$\log_{3/2}\left(\dfrac32\right)^x = \log_{3/2}(3) \\\\ \implies x \log_{3/2}\left(\dfrac 32\right) = \log_{3/2}(3) \\\\ \implies \boxed{x = \log_{3/2}(3)}$

Alternatively, you can divide both sides by $3^x$ :

$\dfrac{3^x}{3^x} = \dfrac{3\cdot 2^x}{3^x} \\\\ \implies 1 = 3 \cdot\left(\dfrac23\right)^x \\\\ \implies \left(\dfrac23\right)^x = \dfrac13$

Then take the base-2/3 logarith of both sides to get

$\log_{2/3}\left(2/3\right)^x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x \log_{2/3}\left(\dfrac23\right) = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(3^{-1}\right) \\\\ \implies \boxed{x = -\log_{2/3}(3)}$

(Both answers are equivalent)

8 0

2 years ago

The area of a compact disc is 452 4/7 square centimeters. What is the diameter of a compact disc ? Use 22/7 as an approximation

motikmotik

$\bf \textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\ \cline{1-1} A=452\frac{4}{7} \end{cases}\implies 452\frac{4}{7}=\pi r^2\implies \cfrac{3168}{7}=\pi r^2 \implies \cfrac{3168}{7\pi }=r^2 \\\\\\ \stackrel{\pi =\frac{22}{7}}{\cfrac{3168}{7\cdot \frac{22}{7}}}=r^2\implies \cfrac{3168}{22}=r^2\implies 144=r^2\implies \sqrt{144}=r\implies 12=r \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{diameter=2r}{d=24}~\hfill$