8-3x=8 solve for X what is X?

Mathematics

1 answer:

NemiM [27]4 years ago

3 0

Answer:

X = 0

Step-by-step explanation:

In the photo!!!

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Can someone PLZZZZ HELP!!!!

swat32

Answer:

The solution is:

Part A. $\sqrt{5}^{\frac{7k}{3}})$ which is sqrt(5)^7k/3[/tex]

Part B. k = 18/7

Step-by-step explanation:

Part A.

To solve this part, we're going two use THREE important properties of exponents:

1. $(x^{n})^{m} = x^{nm}$

2. $\frac{x^{n}}{x^{m}} = x^{n-m}$

3. $\sqrt[n]{x^{m}} = x^{\frac{m}{n} }$

Let's work the numerator using the properties 1, 2 and 3:

$(\sqrt{5}^{3} )^{\frac{k}{9} } } = (\sqrt{5}^{3\frac{k}{9}}) = (\sqrt{5}^{\frac{k}{3}})$

Let's work the denominator using the properties 1, 2 and 3:

$(\sqrt{5}^{6} )^{-\frac{k}{3} } } = (\sqrt{5}^{6\frac{k}{3}}) = (\sqrt{5}^(2k))$

Now dividing the numerator by the denominator:

$\sqrt{5}^{\frac{k}{3}-(-2k)})=\sqrt{5}^{\frac{7k}{3}})$

Part B

if $5^{\frac{3}{2} } 5^{\frac{3}{2}} = \sqrt{5}^{\frac{7k}{3}})$

Then:

$5^{3} = 5^{\frac{7k}{6}})$

So $\frac{7k}{6}} = 3$

Solving for k, we have:

k = 18/7

7 0

4 years ago

The length of a line segment is 7. Its end points are (1, 3) and (k, 3). Solve for k. Is there more than one solution? Explain.

Brilliant_brown [7]

$Endpoints:\\\\A(1,3)\ and\ B(k,3)\\\\Formula\ for\ length\ of\ line:\\\\
|AB|=\sqrt{(x_b-x_a)^2+(y_b-y_a)^2}\\\\
|AB|=7\\\\7=\sqrt{(k-1)^2+(3-3)^2}\\\\
7=\sqrt{k^2-2k-1}\ \ |^2\\\\
49=k^2-2k-1
$ $49=k^2-2k-1\\\\
-k^2+2k+1+49=0\\\\
-k^2+2k+50=0\\\\ \Delta=b^2-4ac\\\\a=-1,\ b=2,\ c=50 \\\\\Delta=(2)^2-4*(-1)*50=4-=4+200=204\\\\ \sqrt{\Delta}=2\sqrt{51}$ $\\\\k_1=\frac{-b-\sqrt{\Delta}}{2a}\ \ k_1=\frac{-2-2\sqrt{51}}{2*(-1)}=1+\sqrt{51} \\\\\ or\ k_2=\frac{-b+\sqrt{\Delta}}{2a}\ \ k_1=\frac{-2+2\sqrt{51}}{2*(-1)}=1-\sqrt{51}$