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Bond [772]
3 years ago
11

A box is formed by cutting squares from the corners of a rectangular piece of cardboard. The volume of the box is found by the e

quation V(h)=h(16-2h)(10-2h) where his a side of the square cutout. Which is the reasonable domain for this situation?
a. (0,5)
b. (0,10)
c. (0,8)
d. (0,infinity)
Mathematics
1 answer:
sineoko [7]3 years ago
5 0

Answer:

A. (0, 5)

Step-by-step explanation:

Physically speaking, the volume of the box and the side of the square cutout are positive value. So that the following inequations must be satisfied:

h > 0 (1)

16-2\cdot h >0 (2)

10-2\cdot h > 0 (3)

From (1), (2), (3) we find that:

(1):

h > 0

(2):

16 > 2\cdot h

8 > h

h < 8

(3):

10 > 2\cdot h

5 > h

h < 5

We see that equation above is the multiplication of a first grade monomial and two first grade binomials, which means that:

h > 0 \,\land\,h < 5\,\land\,h< 8

0< h < 5

In other words, the reasonable domain for this situation is A.

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Answer:

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Step-by-step explanation:

Hope this helps you:)

5 0
3 years ago
PLS DO THIS AS QUICK AS POSSIBLE
VladimirAG [237]

Answer:

a) 1.892a

b.) 5.1x - 1.49y

1.) X = 0.36

2.) Y = 4.347

Car B went faster by 1.25 times

Step-by-step explanation:

a)2.15a+0.03∙(3.4a−12a)

Open the bracket

2.15a + 0.102a - 0.36a

2.252a - 0.36a

1.892a

b)3.27x+1.28y+1.83x−2.77y

Rearranged the algebraic expression above

3.27x + 1.83x + 1.28y - 2.77y

5.1x - 1.49y

1)22.5x−4.6=3.5

Collect the like term

22.5x = 8.1

X = 0.36

2)y÷2.3+5.69=7.58

Y/2.3 = 7.58 - 5.69

Y = 2.3 × 1.89

Y = 4.347

WORD PROBLEM: Car A went 60 km in 5/6 hours while car B went 54 km in 2/3 hours.

Which car was faster?

That's speed

Speed = distance/time

Speed A = 60 ÷ 5/6 = 72 km/h

Speed B = 54 ÷ 2/3 = 81 km/h

How many times faster? 

5/6 ÷2/3

5/6 × 3/2

15/12

1.25 seconds faster

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