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Shtirlitz [24]
2 years ago
5

Hi guys, how to calculate this in system? {2a+3b=14 {5b+4a=20 Thanks,

Mathematics
1 answer:
Tom [10]2 years ago
5 0

Step-by-step explanation:

there it is

and if u need more help tell me

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A piece of cardboard has the dimensions (x + 15) inches by (x) inches with the area of 60 in 2 . Write the quadratic equation th
pochemuha

Answer:

x^2 + 15x - 60 = 0

The actual dimension is 18.28 by 3.28

Step-by-step explanation:

Given

Dimension:

(x + 15)\ by\ x

Area = 60in^2

Required

Determine the quadratic equation and get the possible values of x

Solving (a): Quadratic Equation.

The cardboard is rectangular in shape.

Hence, Area is calculated as thus:

Area = Length * Width

60= (x + 15) * x

Open Bracket

60= x^2 + 15x

Subtract 60 from both sides

x^2 + 15x - 60 = 0

<em>Hence, the above represents the quadratic equation</em>

Solving (b): The actual dimension

First, we need to solve for x

This can be solved using quadratic formula:

x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}

Where

a = 1

b = 15

c = -60

So:

x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}

x = \frac{-15 \± \sqrt{15^2 - 4*1*-60}}{2*1}

x = \frac{-15 \± \sqrt{225 + 240}}{2}

x = \frac{-15 \± \sqrt{465}}{2}

x = \frac{-15 \± \21.56}{2}

Split:

x = \frac{-15 + 21.56}{2} or x = \frac{-15 - 21.56}{2}

x = \frac{6.56}{2} or x = \frac{-36.36}{2}

x = 3.28 or x = -18.18

But length can't be negative;

So:

x = 3.28

The actual dimensions: (x + 15)\ by\ x is

Length =3.28 +15

Length =18.28

Width = x

Width =3.28

<em>The actual dimension is 18.28 by 3.28</em>

3 0
3 years ago
Suppose y varies directly with x. If y = 6 when x = 2, find x when y = 12. 4 3
kobusy [5.1K]

\bf \qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad \stackrel{\textit{constant of variation}}{y=\stackrel{\downarrow }{k}x~\hfill } \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \textit{we also know that } \begin{cases} y=6\\ x=2 \end{cases}\implies 6=k(2)\implies \cfrac{6}{2}=k\implies 3=k \\\\\\ therefore\qquad \boxed{y=3x} \\\\\\ \textit{when y = 12, what is \underline{x}?}\qquad 12=3x\implies \cfrac{12}{3}=x\implies 4=x

6 0
3 years ago
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