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BabaBlast [244]
2 years ago
6

4.A counterexample can be used to show that a conjecture is false. Below are three

Mathematics
1 answer:
Fynjy0 [20]2 years ago
8 0

Answer:

Step-by-step explanation:

1. In some zoos the are more elephants than giraffes, then this conjecture can be true sometimes, however, this can not always be true because some zoo could have more giraffes

2. This can be true unless that turtle ages, or another is found that is taken into captivi9ty and found to be older

3. While most American bison are usaually below 6 feet, that is no gureantee there is no occasaional taller American Bison.

I do not know if these meet your teacher's starndard, but I hope they do. Please thanks and rate!

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Given that (5,-4) is on the graph of f(x) find the corresponding point for the function -1/4f(x)
dusya [7]

For (x, f(x)) = (5, -4), you want to find (x, -1/4·f(x)). That will be

... (5, -1/4·(-4)) = (5, 1)

7 0
4 years ago
Simplify the expression.<br> (X^2/5)^10
Rashid [163]
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3 0
3 years ago
a rectangle has a width that is 7 centimeters less than its length, and it’s area is 330 square centimeters. what are the dimens
WINSTONCH [101]

Answer:

The dimensions of the rectangle are 22cm of length and 15cm of width

Step-by-step explanation:

To solve this we first have to know the formula to calculate the area of ​​a rectangle

a = area = 330 cm²

L = length  =

w = width = L - 7cm

a = l * w

we replace with the known values

330 cm² = L * (l - 7cm)

330 cm² = L² - 7Lcm

0 = L² - 7Lcm - 330 cm²

when we have an equation like this we can use bhaskara

a = 1

b = -7Lcm

c = -330cm²

ax² + bx + c = 0

x = -b(±)√(b² - 4ac)/2a

we replace with the known values

L = -(-7cm)(±)√(7² - 4(1)(-330cm²)) / 2(1)

L = 7cm(±)√(49 + 1320cm²)) / 2

L = 7cm(±)√(1369cm²)) / 2

L1 = (7cm + 37cm) / 2

L1 = 44cm / 2 = 22cm

L2 = (7cm - 37cm) / 2

L2 = -30cm / 2 = -15cm

The positive represents the unknown with which we work (L) and the negative with which we do not work (W)

The dimensions of the rectangle are 22cm of length and 15cm of width

6 0
3 years ago
In the figure below,
almond37 [142]
The answer would be 42. this being because of the fact that 102 can be reflected. you were supplied with the 60 and you want MLO soo subtract 60 from 102 thus giving you 42
3 0
3 years ago
Write an indirect proof..if two angles are supplementary, than they both cannot be obtuse angles.
Aleksandr-060686 [28]
Let's assume that both angles are, in fact, obtuse supplements. We know that supplementary angles must add up to 180°. We also know by definition that obtuse angles are greater than 90°. If we were to take the two supplementary, obtuse angles, ∡A=90+x and ∡B=90+y, with x and y equaling positive real numbers, then we should be able to say that 180=m∡A+mB, or 180=90+x+90+y. By simplifying we get that 180=180+x+y. Simplify further and you get that 0=x+y. If we define x in terms of y, then x= -y. If we define y in terms of x, then y= -x. Because either x or y must be negative to make this statement true, one of the angle measures must be less than 90. If one of the angles must be less than 90 while the other is greater than 90, then one angle MUST be acute if the other is obtuse in order for them to be supplements of each other. 
4 0
3 years ago
Read 2 more answers
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