How do you write y-3=5(4-x) in standard form

Mathematics

1 answer:

Mazyrski [523]2 years ago

7 0

Answer:

y - 3 = 5 (4 - x) written in standard form is 5x + y = 23

Hope this helps!

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7-(-12)=7+(-12)=-5 describe and correct the error in finding the difference 7-(-12)

valina [46]

Answer:

7 - (-12) = 7 + 12 = 19

Step-by-step explanation:

- (-12) is a double negative which means that the minus signs cancel out causing it to to be a positive 12

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How can I find the exact distance between the points(√7,-√2)and (4√7,5√2)

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$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ \sqrt{7}}}\quad ,&{{ -\sqrt{2}}})\quad 
% (c,d)
&({{4\sqrt{7}}}\quad ,&{{ 5\sqrt{2}}})
\end{array}~~~
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
$

$\bf d=\sqrt{[4\sqrt{7}-\sqrt{7}]^2~+~[5\sqrt{2}-(-\sqrt{2})]^2}
\\\\\\
d=\sqrt{(4\sqrt{7}-\sqrt{7})^2~+~(5\sqrt{2}+\sqrt{2})^2}\implies d=\sqrt{(3\sqrt{7})^2~+~(6\sqrt{2})^2}
\\\\\\
d=\sqrt{3^2\cdot 7~~+~~6^2\cdot 2}\implies d=\sqrt{63+72}\implies d=\sqrt{135}
\\\\\\
\begin{cases}
135=5\cdot 3\cdot 3\cdot 3\\
\qquad 5\cdot 3^2\cdot 3\\
\qquad 15\cdot 3^2
\end{cases}\implies d=\sqrt{15\cdot 3^2}\implies d=3\sqrt{15}$

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