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MAVERICK [17]
2 years ago
10

Where does the circumference of each kind of triangle lie? Question 1 options: Inside the triangle. Outside the triangle. On the

hypotenuse. 1. Acute Triangle 2. Obtuse Triangle 3. Right Triangle

Mathematics
1 answer:
stiks02 [169]2 years ago
4 0

Answer:

Inside the triangle - Acute Triangle

Outside the triangle - Obtuse Triangle

On the hypotenuse - Right Triangle

Step-by-step explanation:

The <u>circumcenter</u> is the point where the perpendicular bisectors of a triangle intersect.

In the special case of a right triangle, the circumcenter lies exactly at the midpoint of the hypotenuse.

The circumcenter of an acute triangle lies inside the triangle.

The circumcenter of an obtuse triangle lies outside the triangle.

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Identify the two whole numbers between which the product lies 6*4/5
jenyasd209 [6]
6*4=24/5=4.8

The two numbers are 4 and 5.
5 0
3 years ago
Read each problem carefully and solve as required then answer the questions that follow
zepelin [54]

Answer:

Step-by-step explanation:

1) Number of children = c

Number of adults = a

Total ticket = 250

c + a = 250 -------------------(I)

Cost of 'c' children tickets = 200c

Cost of 'a' adult tickets =  450a

Total cost = 76000

200c + 450a = 76000  ------------------(II)

Multiply (I) by (-200) and then add

(I)*(-200)          -200c - 200a = -50000

(II)                    <u> 200c + 450a = 76000 </u>

<u> </u>                                    250a = 26000

                                           a = 26000/250

a = 104

Plug in a = 104 in equation (1)

c + 104 = 250

         c = 250 - 104

        c = 146

2)Number of children = 146

Number of adults = 104

7 0
3 years ago
Solve<br> x2 - 8x + 14 = 2x - 7
BabaBlast [244]

Answer:

x=7, x=3

Step-by-step explanation:

             x^{2} -8x+14=2x-7

x^{2} -8x-2x+14+7=0

            x^{2} -10x+21=0

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}

x= \frac{-b\pm \sqrt{b^2-4ac}}{2a}

positive

x=\frac{-\left(-10\right)+\sqrt{\left(-10\right)^2-4\cdot \:1\cdot \:21}}{2\cdot \:1}

x=7

negative

x=\frac{-\left(-10\right)-\sqrt{\left(-10\right)^2-4\cdot \:1\cdot \:21}}{2\cdot \:1}

x=3

7 0
3 years ago
Help me pleaseeeeeeeeeeeee
Phoenix [80]
The answer is B :))))
6 0
3 years ago
Read 2 more answers
HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!<br> Show your work
cestrela7 [59]
Well start with reflecting abc over the y axis
7 0
3 years ago
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