The three sides 9 in, 12 in, and 15 in do represent a right triangle. Since the square of the hypotenuse is equal to the sum of the squares of the other two sides, this is a right triangle

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3 years ago

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HEEEEEELP<br><br> question in the attached picture

ryzh [129]

1) $x^{2} +2x-10=0$
a=1
b=2
c=-10
$x= \frac{-(2) +or- \sqrt{ 2^{2}+4(1)(-10) } }{2(1)}$
$x= \frac{-2 + or - \sqrt{4+40} }{2}$
$x= \frac{-2 +or- \sqrt{44} }{2}$
$x= \frac{-2+or- 2\sqrt{11} }{2}$
$x= -1+or- \sqrt{11}$
$\sqrt{11}=3.32$
-1+3.32 = 2.32
-1-3.32 = -4.32

2) $x^{2} -2x-9=0
$
a=1
b=-2
c=-9
$x= \frac{-(-2)+or- \sqrt{ (-2)^{2}+4(1)(-9) } }{2(1)}$
$x= \frac{2+or- \sqrt{4+36} }{2}$
$x= \frac{2+or- \sqrt{40} }{2}$
$x= \frac{2+or-2 \sqrt{10} }{2}$
$x=1+or- \sqrt{10}$
$\sqrt{10} =3.16$
1+3.16 = 4.16
1-3.16 = -2.16

3) $x^{2} -9x+5=0$
a=1
b=-9
c=5
$x= \frac{-(-9)+or- \sqrt{(-9)^{2}-4(1)(5) } }{2}$
$x= \frac{9+or- \sqrt{81-20} }{2}$
$x= \frac{9+or- \sqrt{61} }{2}$
$x= \frac{9}{2} +or- \frac{ \sqrt{61} }{2}$
$\sqrt{61} = 7.81$
$\frac{7.81}{2} =3.91$
$\frac{9}{2}=4.5$
4.5+3.91 = 8.41
4.5-3.91 = 0.59

6 0

4 years ago

10/8c - 4/8 = 6/8 + 5/8c​

10/8c - 4/8 = 6/8 + 5/8c