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3 years ago

What is the equation of this graph

Mathematics

1 answer:

son4ous [18]3 years ago

4 0

Answer:

y=-(2/3) + 1

Step-by-step explanation:

-(2/3) is slope

1 is y intercept

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A truck is being filled with cube-shaped packages that have side lengths of 1/4 foot. The part of the truck that is being filled

n200080 [17]

Answer:

24000 pieces.

Step-by-step explanation:

Given:

Side lengths of cube = $\frac{1}{4} \ foot$

The part of the truck that is being filled is in the shape of a rectangular prism with dimensions of 8 ft x 6 1/4 ft x 7 1/2 ft.

Question asked:

What is the greatest number of packages that can fit in the truck?

Solution:

First of all we will find volume of cube, then volume of rectangular prism and then simply divide the volume of prism by volume of cube to find the greatest number of packages that can fit in the truck.

$Volume\ of\ cube =a^{3}$

$=\frac{1}{4} \times\frac{1}{4}\times \frac{1}{4} =\frac{1}{64} \ cubic \ foot$

Length = 8 foot, Breadth = $6\frac{1}{4} =\frac{25}{4} \ foot$ , Height = $7\frac{1}{2} =\frac{15}{2} \ foot$

$Volume\ of\ rectangular\ prism =length\times breadth\times height$

$=8\times\frac{25}{4} \times\frac{15}{2} \\=\frac{3000}{8} =375\ cubic\ foot$

The greatest number of packages that can fit in the truck = Volume of prism divided by volume of cube

The greatest number of packages that can fit in the truck = $\frac{375}{\frac{1}{64} } =375\times64=24000\ pieces\ of\ cube$

Thus, the greatest number of packages that can fit in the truck is 24000 pieces.

7 0

3 years ago

in Hanley's school, 0.723 of the students participated in an event. if there are 1000 students in Hanley School, how many partic

Ugo [173]

Answer:

723 students participated in the event

Explanations:

The total number of students in Hanley's school = 1000

0.723 of the students participated in an event

Number of students that participated in the event = 0.723 x 1000

Number of students that participated in the event = 723

4 0

1 year ago

Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer

agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming: the function is $f(x)=x^{2}$ in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

$g'(0)=g'(1)$

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

$g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0$

$g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2$

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

$f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]$