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3 years ago

What are the solutions of the quadratic equation (x+3)2=49?

2 answers:

8 0

$\bf (x+3)^2=49\implies x+3=\pm\sqrt{49}\implies x+3=\pm 7
\\\\\\
x=\pm 7 -3\implies x=
\begin{cases}
4\\
-10
\end{cases}$

vredina [299]3 years ago

7 0

Answer:

its C on edge

Step-by-step explanation:

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The diagonals of a rhombus intersect at the point (0,4). If one endpoint of the longer diagonal is located at point (4,10), wher

Wewaii [24]

Answer:

The other endpoint is located at (-4,-2)

Step-by-step explanation:

we know that

The diagonals of a rhombus bisect each other

That means-----> The diagonals of a rhombus intersect at the midpoint of each diagonal

The point (0,4) is the midpoint of the two diagonals

The formula to calculate the midpoint between two points is equal to

$M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

we have

$M=(0,4)$

$(x_1,y_1)=(4,10)$

substitute

$(0,4)=(\frac{4+x_2}{2},\frac{10+y_2}{2})$

<em>Find the x-coordinate $x_2$ of the other endpoint</em>

$0=\frac{4+x_2}{2}$

$x_2=-4$

<em>Find the y-coordinate $y_2$ of the other endpoint</em>

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$8=10+y_2$

$y_2=-2$

therefore

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