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Nataliya [291]
2 years ago
11

What is the distance between (5,2) and (-3,4) on a coordinate plane

Mathematics
1 answer:
maksim [4K]2 years ago
5 0
\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ 5}}\quad ,&{{ 2}})\quad 
%  (c,d)
&({{ -3}}\quad ,&{{ 4}})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
d=\sqrt{[-3-5]^2+[4-2]^2}
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Solve the equation.<br><br> 3(x−9)=30
a_sh-v [17]

Answer:

x = 19

Step-by-step explanation:

3(x - 9) = 30

Divide both sides by 3 to isolate the binomial.

x - 9 = 10

Add 9 to both sides to isolate x.

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Check your answer by plugging x = 19 back into the equation.

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Subtract.

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Multiply.

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Your answer is correct.

Hope this helps!

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2 years ago
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<img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B80%7D%20" id="TexFormula1" title=" \sqrt[3]{80} " alt=" \sqrt[3]{80} " ali
Marianna [84]

Start by decomposing the number inside the root into primes

Then group the terms into cubes if possible

\begin{gathered} 80=2\cdot2\cdot2\cdot2\cdot5 \\ 80=2^3\cdot2\cdot5 \\ 80=10\cdot2^3 \end{gathered}

rewrite the root

\sqrt[3]{80}=\sqrt[3]{10\cdot2^3}

then cancel the terms that are cubes and bring them out of the root

\sqrt[3]{80}=2\sqrt[3]{10}

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The answer to the question

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