Select the two values of x that are roots of this equation. 2x2 + 11x - 6 = 0

Mathematics

1 answer:

sukhopar [10]3 years ago

8 0

Answer:

2×2 +11x - 6 =0

4 + 11x - 6 = 0

11x - 6 =4

-6 + 4 = 11x

-2 = 11x

X =11/2

X =5.2

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Need help with 7A plz

Eduardwww [97]

A) area of square = base*height/////Triangle= 0.5*base*height

=(a^1/3*b^3/4)*(a^2/3*b^1/2)+(1/2*a^1/3*b^3/4*a^2/3*b^1/4)
=(a^1*b^5/4)+(1/2*a^1*b^1)
=(a*b^5/4)+(1/2*a*b)

B) a^2+b^2=c^2 (pythagorean theorem)
((27^2/3)*(16^1/4))^2 + (27^1/3)*(16^3/4)
(9*2=(18^2))=324 + (3*8= 24^2)= 576
324+576= 900
(900)^1/2= 30
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C)( a^2/3*b^1/2)= 36*2(two sides)= 72
(a^1/3b^3/4)=24
(a^2/3*b^1/4)= 18
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4 years ago

51: Y = 3: 5 value of Y

ivann1987 [24]

Answer:Y=25:3

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A fraction becomes 4÷5 if 1 is added to both numerator and denominator. If, however, 5 is subtracted from both numerator and den

tia_tia [17]

Answer:

$\dfrac{7}{9}$

Step-by-step explanation:

$\dfrac{x+1}{y+1}=\dfrac{4}{5}\\\Rightarrow 5x-4y=-1$

$\dfrac{x-5}{y-5}=\dfrac{1}{2}\\\Rightarrow 2x-y=5$

Putting it in matrix form

$\begin{bmatrix}a_{1}&b_{1}\\a_{2}&b_{2}\end{bmatrix}{\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}{c_{1}}\\{c_{2}}\end{bmatrix}\\\Rightarrow\begin{bmatrix}5 & -4\\2 & -1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}-1\\ 5\end{bmatrix}$

From Cramer's rule we have

$x=\dfrac{\begin{vmatrix}c_1 &b_1 \\ c_2 & b_2\end{vmatrix}}{\begin{vmatrix}a_1 &b_1 \\ a_2& b_2\end{vmatrix}}\\\Rightarrow x=\dfrac{\begin{vmatrix}-1 &-4 \\ 5 & -1\end{vmatrix}}{\begin{vmatrix}5 &-4 \\ 2& -1\end{vmatrix}}\\\Rightarrow x=\dfrac{1+20}{-5+8}\\\Rightarrow x=7$

$y=\dfrac{\begin{vmatrix}a_1 &c_1 \\ a_2 & c_1\end{vmatrix}}{\begin{vmatrix}a_1 &b_1 \\ a_2& b_2\end{vmatrix}}\\\Rightarrow y=\dfrac{\begin{vmatrix}5 &-1 \\ 2 & 5\end{vmatrix}}{\begin{vmatrix}5 &-4 \\ 2& -1 \end{vmatrix}}\\\Rightarrow y=\dfrac{25+2}{-5+8}\\\Rightarrow y=9$