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maw [93]
3 years ago
15

HELP PLEASE

Mathematics
2 answers:
marin [14]3 years ago
6 0
Scalene - the sides are all different sizes 
acute - all the angles are lass than 90 degrees

sweet [91]3 years ago
5 0
This is an isosceles acute triangle.
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Draw Three triangles One equilateral one isosceles and one scalene label Each and explain how you classified each triangle
Fantom [35]
Here is a picture of each.

Here are the definitions of each:
1. Equilateral- this triangle has side lengths that are all the same (congruent).
2. Isosceles-this triangle has exactly 2 sides that are the same length(congruent).
3. Scalene- this triangle has no sides that are the same length.

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3 years ago
Receipt-of-goods discounts tend to be offered in cases where?
AleksAgata [21]
<span>The invoice is enclosed with the goodsand arrives at the same time that they do</span>
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Read 2 more answers
Sorry I have one more question, 23 hundredths more than 9.041
marissa [1.9K]
23 hundredths = 0.23

9.041 + 0.23 = 9.271
5 0
3 years ago
3) What are the new coordinates of A’, B’, and C’ if triangle ABC is rotated 270 degrees clockwise? Write the transformation rul
MissTica

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3 0
3 years ago
What is the maximum vertical distance between the line
astra-53 [7]

Answer: the maximum distance is \frac{289}{4} and can be found at x = \frac{1}{2}


Let's call:

f(x) = x + 72

g(x) = x²


A point belonging to the line will be L(x, x+72) and a point on the parabola will be P(x, x²). It can be easily seen that in the interval -8 ≤ x ≤ 9 the line is above the parabola (it's enough to graph them or plug in some numbers), therefore their distance at any point will be:

d(x) = f(x) - g(x) = - x² + x + 72


The function d(x) is a parabola that opens downward, therefore the maximum will be the vertex; given a parabola

y(x) = ax² + bx + c

the coordinates of the vertex will be

V(\frac{-b}{2a}, y(\frac{-b}{2a}))


Therefore:

V(\frac{1}{2}, \frac{289}{4})


Hence, the maximum distance is \frac{289}{4} = 72.25 and can be found at x = \frac{1}{2}


Another way to find the maximum is to use calculus to find the first derivative of the distance:

d'(x) = -2x + 1


and set it equal to zero:

-2x + 1 = 0

x = \frac{1}{2}


Since the second derivative:

d"(x) = -2 

is negative, the point is a maximum.


Then, substitute this value in the equation for the distance:

d(\frac{1}{2}) = -(\frac{1}{2})^{2} + \frac{1}{2} + 72

d(\frac{1}{2}) = \frac{289}{4} = 72.25


<span>Hence, the maximum distance is </span>\frac{289}{4} = 72.25 and can be found at x = \frac{1}{2}

7 0
3 years ago
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