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laiz [17]
3 years ago
9

Really need help!!! quadratic equations UGH

Mathematics
2 answers:
Mumz [18]3 years ago
5 0
2m³ + 10 m = 48

2m² + 10m - 48 = 0, solve this quadratic for m. Apply the formula:

m'=[-b + √(b²+4ac)]/2a and m" ==[-b - √(b²+4ac)]/2a

WHERE  a = 2, b=10, c= - 48
and you will find :
m=  3 and m= - 8 (answer a)
kati45 [8]3 years ago
4 0
2m^2 + 10m = 48\\
2m^2+10m-48=0\\
m^2+5m-24=0\\
m^2-3m+8m-24=0\\
m(m-3)+8(m-3)=0\\(m+8)(m-3)=0\\
m=-8 \vee m=3
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Find the constant of proportionality k. Then write an equation for the relationship between x and y
ExtremeBDS [4]

Question:

Find the constant of proportionality k. Then write an equation for the relationship between x and y

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Answer:

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Step-by-step explanation:

Given

\begin{array}{ccccc}x & {2} & {4} & {6} & {8} \ \\ y & {10} & {20} & {30} & {40} \ \ \end{array}

Solving (a): The constant of proportionality:

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The constant of proportionality k is:

k = \frac{y_2 - y_1}{x_2 - x_1}

k = \frac{30-10}{6-2}

k = \frac{20}{4}

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Solving (b): The equation

In (a), we have:

(x_1,y_1) = (2,10)

k can also be expressed as:

k = \frac{y- y_1}{x- x_1}

Substitute values for x1, y1 and k

5 = \frac{y- 10}{x- 2}

Cross multiply:

y - 10 = 5(x - 2)

Open bracket

y - 10 = 5x - 10

Add 10 to both sides

y - 10 +10= 5x - 10+10

y = 5x

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