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NeX [460]
3 years ago
6

A block of cheese is in the shape of a rectangle prism with a width of 2.5 inches in the height of 5 inches the volume of the bl

ock of cheese is 75 cubic inches what is the length of the block of cheese
Mathematics
2 answers:
zzz [600]3 years ago
8 0
V=xyz

75=2.5(5)L

L=75/12.5

L=6in
brilliants [131]3 years ago
5 0
The formula for volume is l x w x h.

To find the missing link, we do the equation backwards.

l x 2.5 x 5 = 75

If we divide 75 by 5 then by 2.5, we get 6 which becomes our length.

To check, do 5 x 2.5 x 6 which should equal 75 

6 is your length
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The magnitude of the vector to two decimal places is 11.40 units

The formula for calculating the magnitude of the vector is expressed as:

D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2 }

Given the coordinate points  (-3, 4) and (6, -3)

D=\sqrt{(-3-4)^2+(6+3)^2 }\\D=\sqrt{(-7)^2+9^2}\\D=\sqrt{49+81}\\D=\sqrt{130}\\D=  11.40

Hence the magnitude of the vector to two decimal places is 11.40 units

Learn more here: brainly.com/question/20759497

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r-ruslan [8.4K]

20. A rational function \frac{p(x)}{q(x)} (where p,q are polynomials in x) has a removable discontinuity at x=a if for some positive integer n we can factorize

p(x) = (x-a)^n p^*(x)

q(x) = (x-a)^n q^*(x)

\implies \dfrac{p(x)}{q(x)} = \dfrac{(x-a)^n p^*(x)}{(x-a)^n q^*(x)} = \dfrac{p^*(x)}{q^*(x)}

That is, we "remove" the discontinuity at x=a because while x\neq a, we have \frac{x-a}{x-a}=1. The discontinuity is still there, since \frac{p(a)}{q(a)} is undefined, but in the limit sense we can essentially ignore the factors of x-a. The graph of \frac{p(x)}{q(x)} will have all the features of \frac{p^*(x)}{q^*(x)} aside from a hole at x=a.

In this case, we have

f(x) = \dfrac{x+4}{x^2-x-20} = \dfrac{x+4}{(x+4)(x-5)}

and when x\neq-4, we can cancel the factor of x+4 so that

f(x) = \begin{cases}\dfrac1{x-5} & \text{if }x \neq -4 \\\\ \text{DNE} & \text{if }x = -4\end{cases}

("DNE" = "does not exist", i.e. is undefined. For some reason "undefined" wouldn't render...)

This means x=-4 is a removable discontinuity.

We cannot do the same with the factor of x-5, so in contrast x=5 is a non-removable discontinuity.

21. The pieces of f(x) defined on x and x>2 are themselves continuous since they are polynomials. Then the continuity of f(x) over the entire real line depends on the point where the pieces meet.

Here we have

f(x) = \begin{cases}x+3 & \text{if }x\le2 \\ cx+6 & \text{if }x>2\end{cases}

so the pieces meet at x=2. Continuity at this point requires that the both limits from either side of x=2 be the same. This means

\displaystyle \lim_{x\to2^-} f(x) = \lim_{x\to2} (x+3) = 2+3 = 5

\displaystyle \lim_{x\to2^+} f(x) = \lim_{x\to2} (cx+6) = 2c + 6

Solve for c.

2c + 6 = 5 \implies 2c = -1 \implies \boxed{c = -\dfrac12}

4 0
2 years ago
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