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Natalka [10]
3 years ago
6

4. Decide if the segment lengths form a triangle. If they do, Indicate whether the triangle is acute, right, or obtuse.

Mathematics
1 answer:
beks73 [17]3 years ago
7 0
To see if line segments may form a triangle basically one must check if the sum of two sides are greater than the third. 
 a + b > c
a + c > b
b + c > a

a. can form a triangle
b. can form a triangle
c. can form a triangle
d. cannot form a triangle as 14 + 21 = 35 which is not greater than 36

a = right triangle with the right angle between 16 and 30
b = acute triangle
c = obtuse triangle with teh obtuse angle between 7 and 24
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Help me it's algebra and dont tell me to look at the lesson :)
suter [353]

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He's supposed to use a total of between 70 and 74 beads, so

70 ≤ b + g ≤ 74

The ratio of green beads to blue beads is g/b, and this ratio has to be between 1.4 and 1.6, so

1.4 ≤ g/b ≤ 1.6

For completeness, Giovanni must use at least one of either bead color, so it sort of goes without saying that this system must also include the conditions

b ≥ 0

g ≥ 0

(These conditions "go without saying" because they are implied by the others. g/b is a positive number, so either both b and g are positive, or they're both negative. But they must both be positive, because otherwise b + g would be negative. I would argue for including them, though.)

7 0
2 years ago
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Helen [10]

Answer: 29/28

Step-by-step explanation:

-2/7 x (-3 5/8)

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= 2/7 x 29/8 = 1/7 x 29/4

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7 0
3 years ago
If one angle of a trapezoid is a right angle, the greatest number of additional angles a trapezoid might have is Select one: a.
Viktor [21]

Answer:

c.

Step-by-step explanation:

7 0
3 years ago
Expand: -1/3(y-x)<br> Expand: 8(6x-4)
almond37 [142]

Answer:

for the second one

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Step-by-step explanation:

5 0
3 years ago
Write the sentence as an equation.<br> the quotient of a and 399 equals 181 times z
stich3 [128]

Answer:

Step-by-step explanation:

Quotient of a and 399 : \dfrac{a}{399}

181 times z = 181*z = 181z

Equation:

\dfrac{a}{399}=181z

8 0
2 years ago
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