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mestny [16]
3 years ago
15

Determine if its possible to form a triangle using the set of segments with the given measurement. explain please! 12 m, 5 m, 18

.9 m
Mathematics
1 answer:
IRINA_888 [86]3 years ago
6 0

Adding together the shorter two sides results in 17 m.  If these two sides were laid end to end, the sum of their sides (17 m) would be less than the length of the third side.  Thus, NO triangle could be constructed with this set of line segments.

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1. Solve the equation.
Keith_Richards [23]

Answer:

1. -3

2. -12

3. 20

4. -15

5. 6

6. -15w+5

7. 10

8. C<-26

9.p> -40

10. P> 16

Step-by-step explanation:

~ -9p -17 = 10

-9p = 27

P = -3

~ x/4 -5 = -8

x/4 = -3

x = -12

~ p/5 +6 =10

p/5 = 4

p = 20

~ -2(m-30) =-6m

-2m+60=-6m

4m = -60

m = -15

~ 3n +2=8+2n

n = 6

8 0
3 years ago
I need help ASAP due at 11:59
algol [13]

Answer:

Step-by-step explanation:

8 0
3 years ago
Jorge is asked to build a box in the shape of a rectangular prism. The maximum girth of the box is 20 cm. What is the width of t
MariettaO [177]

Answer:

The width of the box is 6.7 cm

The maximum volume is 148.1 cm³

Step-by-step explanation:

The given parameters of the box Jorge is asked to build are;

The maximum girth of the box = 20 cm

The nature of the sides of the box = 2 square sides and 4 rectangular sides

The side length of square side of the box = w

The length of the rectangular side of the box = l

Therefore, we have;

The girth = 2·w + 2·l = 20 cm

∴ w + l = 20/2 = 10

w + l = 10

l = 10 - w

The volume of the box, V = Area of square side × Length of rectangular side

∴ V = w × w × l = w × w × (10 - w)

V = 10·w² - w³

At the maximum volume, we have;

dV/dw = d(10·w² - w³)/dw = 0

∴ d(10·w² - w³)/dw = 2×10·w - 3·w² = 0

2×10·w - 3·w² = 20·w - 3·w² = 0

20·w - 3·w² = 0 at the maximum volume

w·(20 - 3·w) = 0

∴ w = 0 or w = 20/3 = 6.\overline 6

Given that 6.\overline 6 > 0, we have;

At the maximum volume, the width of the block, w = 6.\overline 6 cm ≈ 6.7 cm

The maximum volume, V_{max}, is therefore given when w = 6.\overline 6 cm = 20/3 cm  as follows;

V = 10·w² - w³

V_{max} = 10·(20/3)² - (20/3)³ = 4000/27 = 148.\overline {148}

The maximum volume, V_{max} = 148.\overline {148} cm³ ≈ 148.1 cm³

Using a graphing calculator, also, we have by finding the extremum of the function V = 10·w² - w³, the coordinate of the maximum point is (20/3, 4000/27)

The width of the box is;

6.7 cm

The maximum volume is;

148.1 cm³

5 0
3 years ago
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