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vladimir2022 [97]
3 years ago
12

Find any real solutions for the following quadratic equations.

Mathematics
1 answer:
stealth61 [152]3 years ago
3 0
2x^2 - 8 = 0
2x^2 = 8
x^2 = 8/2 = 4
x = sqrt(4) = 2 or -2
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The area of a rectangle is 216 square inches. If the perimeter is 60 inches, find the length and width of the rectangle
mojhsa [17]

A=Lw

P=2L+2w

216=Lw

W=216/L...substitute to Perimeter equation

60=2L+2(216/L)

60L=2L^2 + 432

2L^2-60L+432=0

2(L^2 - 30L + 216)=0

2(L-18)(L-12)=0

L=12, W=18 or L=18, W=12

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3 years ago
Write the equation of the line in point-slope form. ASAP PLS !!!!
Vlad1618 [11]

Answer:

y - 1 = \frac{2}{3}(x - 3)

Step-by-step explanation:

The equation of a line in point- slope form is

y - b = m(x - a)

where m is the slope and (a, b) a point on the line

Calculate m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (0, - 1) and (x₂, y₂ ) = (3, 1) ← 2 points on the line

m = \frac{1+1}{3-0} = \frac{2}{3}

and using (a, b) = (3, 1 ) , then

y - 1 = \frac{2}{3} (x - 3)

8 0
2 years ago
5x 2(x - 3) = -2(x - 1)
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Solve for x and y matrix
charle [14.2K]

Answer:

x=-1

y=5

Step-by-step explanation:

\begin{pmatrix}x&y\\ \:-1&4\end{pmatrix}\begin{pmatrix}2&5\\ \:4&1\end{pmatrix}=\begin{pmatrix}18&0\\ \:14&-1\end{pmatrix}

\begin{pmatrix}x&y\\ -1&4\end{pmatrix}\begin{pmatrix}2&5\\ 4&1\end{pmatrix}

=\begin{pmatrix}x\times \:2+y\times \:4&5x+y\\ 14&-1\end{pmatrix}

=\begin{pmatrix}18&0\\ 14&-1\end{pmatrix}

\begin{bmatrix}x\times \:2+y\times \:4=18\\ 5x+y=0\end{bmatrix}

x=-1

y=5

<u>OAmalOHopeO</u>

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2 years ago
The equation with one change and no solutions: 8x + 6 =
Lisa [10]
The answer is 2 ........
6 0
3 years ago
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