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3 years ago

Find the quotient: 9.576 divided by 2.1 ❤what is the answer

Mathematics

2 answers:

vlada-n [284]3 years ago

8 0

The answer is 4.56. I hope this helps you

Lilit [14]3 years ago

4 0

9.576 ÷ 2.1 = 4.56

Hope this helps!

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3 years ago

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Furkat [3]

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3 years ago

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Help please, finals week is upon me

kenny6666 [7]

The circumcenter is found by finding the intersection of at least 2 perpendicular bisector segments.

Find the perpendicular bisector to segment AB. This is the line y = -3.5; the idea is that you find the equation of the horizontal line through the midpoint of AB. The midpoint has a y coordinate of -3.5. This line is shown in red horizontal line in the attached image below.

The midpoint of AC is 2.5, so the perpendicular bisector to AC is x = 2.5 which is shown as the vertical green line in the same diagram.

The red and green lines cross at the location (2.5, -3.5) which is the circumcenter's location. If you were to draw a circle through all three points A, B, & C, then this circle would be centered at (2.5, -3.5)

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<h3>Answer: (2.5, -3.5)</h3>

note: 2.5 in fraction form is 5/2 while -3.5 in fraction form is -7/2

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3 years ago

Solve.

Ksenya-84 [330]

<span>Solve.

x² + 4x + 4 = 18

A x=−4±3√ 2

B x=2±3√ 2

C x=4±9√ 2

D x=−2±3√2

</span> $x^2 + 4x + 4 = 18 \\ \\ x^2+4x+4-18= 0 \\ \\ x^2+4x-14=0 \\ \\ x_1_y_2= \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} \qquad a= 1\qquad b= 4\qquad c= -14 \\ \\ \\ x_1_y_2= \dfrac{-4\pm \sqrt{4^2-4(1)(-14)} }{2(1)} \\ \\ \\ x_1_y_2= \dfrac{-4\pm \sqrt{16-(-56)} }{2} \\ \\ \\ x_1_y_2= \dfrac{-4\pm \sqrt{72} }{2} \\ \\ \\ x_1= \dfrac{-4+ \sqrt{72} }{2} \qquad\qquad x_2= \dfrac{-4+ \sqrt{72} }{2}\\ \\ \\ x_1= \dfrac{-4+ \sqrt{2^3*3^2} }{2} \qquad\qquad x_2= \dfrac{-4- \sqrt{2^3*3^2} }{2}$
<span>
</span> $x_1= \dfrac{-4+2*3 \sqrt{2} }{2} \qquad\qquad x_2= \dfrac{-4- 2*3\sqrt{2} }{2} \\ \\ \\ x_1= \dfrac{-4+6 \sqrt{2} }{2} \qquad\qquad x_2= \dfrac{-4- 6\sqrt{2} }{2} \\ \\ \\ x_1= \dfrac{-4}{2} + \dfrac{6 \sqrt{2} }{2} \qquad\qquad x_2= \dfrac{-4}{2}- \dfrac{ 6\sqrt{2} }{2} \\ \\ \\ x_1= -2 + 3 \sqrt{2} \qquad\qquad\quad x_2= -2- 3\sqrt{2} \\ \\ \\ \boxed{x= -2\pm 3\sqrt{2} } \to D)$ <span>
</span>

7 0

4 years ago

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