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3 years ago

Solve this proportion: 8/x = 2/5. x =

Mathematics

1 answer:

Taya2010 [7]3 years ago

3 0

8/x= 2/5

8•5= 2•x

40=2x

40/2= x

20= x

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20*30 + 10*10 = $700 ft^2$

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3 years ago

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3 years ago

Is it <br> a.x = _<br> b. the solution is all real numbers <br> c. there is no solution

Zielflug [23.3K]

The solution to the given algebraic equation 1000(7x - 10) = 50(292 + 100x) is; x = 12.3

<h3>How to simplify algebraic equations?</h3>

We want to simplify the algebraic equation;

1000(7x - 10) = 50(292 + 100x)

The first step is to use distributive property of equality to expand the bracket to get;

7000x - 10000 = 14600 + 5000x

7000x - 5000x = 14600 + 10000

2000x = 24600

Use division property of equality to divide both sides by 2000 to get;

x = 24600/2000

x = 12.3

Thus, we can conclude that the solution to the given algebraic equation 1000(7x - 10) = 50(292 + 100x) is; x = 12.3

Read more about algebraic equations at; brainly.com/question/723406

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1 year ago

Use the four-step definition of the derivative to find f'(x) if f(x) = −4x^3 −1.

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$\stackrel{de finition \textit{ of a derivative as a limit}}{\lim\limits_{h\to 0}~\cfrac{f(x+h)-f(x)}{h}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{[-4(x+h)^3-1]~~ - ~~[-4x^3-1]}{h} \\\\\\ \cfrac{[-4(x^3+3x^2h+3xh^2+h^3)-1]~~ ~~+4x^3+1}{h} \\\\\\ \cfrac{[-4x^3-12x^2h-12xh^2-4h^3-1]~~ ~~+4x^3+1}{h} \\\\\\ \cfrac{-4x^3-12x^2h-12xh^2-4h^3-1+4x^3+1}{h}\implies \cfrac{-12x^2h-12xh^2-4h^3}{h}$

$\cfrac{h(-12x^2-12xh-4h^2)}{h}\implies -12x^2-12xh-4h^2 \\\\\\ \lim\limits_{h\to 0}~-12x^2-12xh-4h^2\implies \lim\limits_{h\to 0}~-12x^2-12x(0)-4(0)^2 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \lim\limits_{h\to 0}~-12x^2~\hfill$

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2 years ago

How many Solutions does this system have? (1 point)

mixas84 [53]

The given system of equation that is $2x+y=3$ and $6x=9-3y$ has infinite number of solutions.

Option -C.

<u>Solution:</u>

Need to determine number of solution given system of equation has.

$\begin{array}{l}{2 x+y=3} \\\\ {6 x=9-3 y}\end{array}$

Let us first bring the equation in standard form for comparison

$\begin{array}{l}{2 x+y-3=0} \\\\ {6 x+3 y-9=0}\end{array}$

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

To check how many solutions are there for system of equations $a_{1} x+b_{1} y+c_{1}=0 \text{ and }a_{2} x+b_{2} y+c_{2}=0$ , we need to compare ratios of $\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}$

In our case,

$a_{1} = 2, b_{1}= 1\text{ and }c_{1}= -3$

$a_{2} = 6, b_{2} = 3,\text{ and }c_{2} = -9$

$\begin{array}{l}{\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{6}=\frac{1}{3}} \\\\ {\Rightarrow \frac{b_{1}}{b_{2}}=\frac{1}{3}} \\\\ {\Rightarrow \frac{c_{1}}{c_{2}}=\frac{-3}{-9}=\frac{1}{3}} \\\\ {\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{3}}\end{array}$

As $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ , so given system of equations have infinite number of solutions.

Hence, we can conclude that system has infinite number of solutions.

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3 years ago