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3 years ago

Approximate square root 18 to the tenths and plot on number line

1 answer:

8 0

Since 18 does not have a number that when multiplied by itself equal 18, then it will be a decimal number. You can tell the closest whole number by seeing what is the closest square. In this case we know that 16 is 4 times 4 and the next one would be 5 times 5 to equal 25, so the whole number would be 4.

Now, the easiest way to do this is to get you calculator and take the square root of 18 to get 4.242641. The problem wants it to be to the nearest tenth so 4.2 is your answer. On a number line, if it is in four parts, you will plot it just before the first little line.

Hope this helps! If you are using more small lines in you number line, comment me below and I would be happy to lead you in the right direction.

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2 years ago

What is the true solution to 3 in 2 + in 8 = 2 in (4x)

11Alexandr11 [23.1K]

The first step for solving this equation is to determine the defined range.
3㏑(2) + ㏑(8) = 2㏑(4x), x > 0
Write the number 8 in exponential form.
3㏑(2) + ㏑(2³) = 2㏑(4x)
Using ㏑( $a^{x}$ ) = x × ㏑(a),, transform the expression.
3㏑(2) + 3㏑(2) = 2㏑(4x)
Now collect the like terms on the left side of the equation.
6㏑(2) = 2㏑(4x)
Switch the sides of the equation.
2㏑(4x) = 6㏑(2)
Using x × ㏑(a) = ㏑( $a^{x}$ ),, transform the expression on the left side of the equation.
㏑((4x)²) = 6㏑(2)
Using x × ㏑(a) = ㏑( $a^{x}$ ),, transform the expression on the right side of the equation.
㏑((4x)²) = ㏑( $2^{6}$ )
Since the bases of the logarithms are the same,, you need to set the arguments equal.
(4x)² = $2^{6}$
Take the square root of both sides of the equation and remember to use both the positive and negative roots.
4x = +/- 8
Now separate the equation into 2 possible cases.
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Solve the top equation for x.
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Solve the bottom equation for x.
x = 2
, x > 0
x = -2
Lastly,, check if the solution is in the defined range to find your final answer.
x = 2
This means that the correct answer to your question is x = 2.
Let me know if you have any further questions
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2 years ago

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What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$

As for infinitely many solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}$

Hence b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

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2 years ago