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3 years ago

.................................

Mathematics

2 answers:

Nutka1998 [239]3 years ago

6 0

{ $x=10$
{ $3*10+5y=20$

{ $x=10$
{ $30+5y=20$

{ $x=10$
{ $5y=-10$

{ $x=10$
{ $y=-2$

The answer: (10;-2)

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Greetings from Poland :)

Alchen [17]3 years ago

4 0

Answer: x = 10, y = -2 ; → Write as: (10, -2) .
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Explanation:
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Given:
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x = 10 ;

3x + 5y = 20 ;
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→ Plug in "10" for the value of "x" in the second equation:
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3*(10) + 5y = 20 ; Solve for "y" ;
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↔ 5y + 3*(10) = 20 ;

5y + 30 = 20 ;
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Subtract "30" from EACH side of the equation:
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5y + 30 − 30 = 20 <span>− 30 ;
</span>
5y = - 10 ;
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→ Divide EACH SIDE of the equation by "5" ; to isolate "y" on one side of the equation; and to solve for "y" ;
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5y / 5 = - 10 / 5 ;
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y = - 2;

And we are given: x = 10 ;
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→ Let us check our answer, by substituting "10" for "x"; and "-2" for "y" ; in the second equation given; to see if the equation holds true:
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Given: ... 3x + 5y = 20 ;

Substitute: 3*(10) + 5*(-2) = ? 20 ?? ;
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30 + (-10) = ? 20 ?? ;
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30 − 10 = ? 20 ? Yes; "30 − 10 = 20 !!! " .
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So, " x = 10 , y = -2 " ; ↔ write as: " (10, -2) " .
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The difference between the roots of the equation 3x^2+bx+10=0 is equal to 4 1/3 . Find b.

telo118 [61]

$\textit{quadratic formula}\\
{{ 3}}x^2{{ +b}}x{{ +10}}=0
\qquad \qquad 
x= \cfrac{ - {{ b}} \pm \sqrt { {{ b}}^2 -4{{ a}}{{ c}}}}{2{{ a}}}
\\ \quad \\
meaning\implies x=\cfrac{-b\pm\sqrt{b^2-4(3)(10)}}{2(3)}\qquad now
\\ \quad \\
\cfrac{-b\pm\sqrt{b^2-4(3)(10)}}{2(3)}\to 
\begin{cases}
\cfrac{-b+\sqrt{b^2-4(3)(10)}}{2(3)}
\\ \quad \\

\cfrac{-b-\sqrt{b^2-4(3)(10)}}{2(3)}
\end{cases}\impliedby \textit{two roots}$

$\textit{and their root difference is }4\frac{1}{3}\implies \cfrac{13}{3}\qquad so
\\ \quad \\
\left[ \cfrac{-b+\sqrt{b^2-4(3)(10)}}{2(3)} \right]-\left[ \cfrac{-b-\sqrt{b^2-4(3)(10)}}{2(3)} \right]=\cfrac{13}{3}
\\ \quad \\
\left[ \cfrac{-b+\sqrt{b^2-4(3)(10)}}{6} \right]+\left[ \cfrac{+b+\sqrt{b^2-4(3)(10)}}{6} \right]=\cfrac{13}{3}
\\ \quad \\
\cfrac{-b+\sqrt{b^2-4(3)(10)}+b+\sqrt{b^2-4(3)(10)}}{6}=\cfrac{13}{3}
\\ \quad \\
\cfrac{2\sqrt{b^2-4(3)(10)}}{6}=\cfrac{13}{3}$

and I'm pretty sure you can take it from there

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fenix001 [56]

Answer:

(8) d. 4/3 (9) a. (X - 4)^2 + (y-4)^2= 25

Step-by-step explanation:

8. CENTER OF CIRCLE: O (2,6)

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if m = 4/3 line: y = 4/3 x

perpendicular distance (d) from center of circle to line y = 4/3 x is

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d = r means line touch with circle

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