HELP PLEASE!!

What is the volume of a cube that has edges of 3 3/4

tino4ka555 [31]

Answer:

3375/64

Step-by-step explanation:

You make the fraction into a improper fraction

3 3/4=15/4

You cube the top and bottom

15^3/4^3=3375/64

Thus

answer-3375/64

8 0

4 years ago

Need help solving problems 29-32 please explain so I can understand concepts thank you!!

Fittoniya [83]

29.<BCA

30. <ACE

31.<ACB

32.<JCB

3 0

3 years ago

Read 2 more answers

Use the definition of continuity to determine whether f is continuous at a.

dmitriy555 [2]

$f(x)$ will be continuous at $x=a=7$ if
(i) $\displaystyle\lim_{x\to7}f(x)$ exists,
(ii) $f(7)$ exists, and
(iii) $\displaystyle\lim_{x\to7}f(x)=f(7)$ .

The second condition is immediate, since $f(7)=8918$ has a finite value. The other two conditions can be established by proving that the limit of the function as $x\to7$ is indeed the value of $f(7)$ . That is, we must prove that for any $\varepsilon>0$ , we can find $\delta>0$ such that

$|x-7|$

Now,

$|f(x)-f(7)|=|5x^4-9x^3+x-8925|$

Notice that when $x=7$ , we have $5x^4-9x^3+x-8925=0$ . By the polynomial remainder theorem, we know that $x-7$ is then a factor of this polynomial. Indeed, we can write

$|5x^4-9x^3+x-8925|=|(x-7)(5x^3+26x^2+182x+1275)|=|x-7||5x^3+26x^2+182x+1275|$

This is the quantity that we do not want exceeding $\varepsilon$ . Suppose we focus our attention on small values $\delta$ . For instance, say we restrict $\delta$ to be no larger than 1, i.e. $\delta\le1$ . Under this condition, we have

$|x-7|$

Now, by the triangle inequality,

$|5x^3+26x^2+182x+1275|\le|5x^3|+|26x^2|+|182x|+|1275|=5|x|^3+26|x|^2+182|x|+1275$

If $|x|$ , then this quantity is moreover bounded such that

$|5x^3+26x^2+182x+1275|\le5\cdot8^3+26\cdot8^2+182\cdot8+1275=6955$

To recap, fixing $\delta\le1$ would force $|x|$ , which makes

$|x-7||5x^3+26x^2+182x+1275|$

and we want this quantity to be smaller than $\varepsilon$ , so

$6955|x-7|$

which suggests that we could set $\delta=\dfrac{\varepsilon}{6955}$ . But if $\varepsilon$ is given such that the above inequality fails for $\delta=\dfrac{\varepsilon}{6955}$ , then we can always fall back on $\delta=1$ , for which we know the inequality will hold. Therefore, we should ultimately choose the smaller of the two, i.e. set $\delta=\min\left\{1,\dfrac{\varepsilon}{6955}\right\}$ .

You would just need to formalize this proof to complete it, but you have all the groundwork laid out above. At any rate, you would end up proving the limit above, and ultimately establish that $f(x)$ is indeed continuous at $x=7$ .

5 0

3 years ago

4. One week Lin's balance at the start of the week was (- $16) but she

Sphinxa [80]

Her balance was $4 at the end of the week

3 0

3 years ago

URGENT!! please help! will mark brainliest!

Crazy boy [7]

Answer:

Step-by-step explanation:

Please refer to the Image attached to this.

The Line passing through the rainbow represents the Airplane Trajectory.

2.

a) Domain and Range :

Domain is the set of values of which a function is satisfied. Range is the set of all possible output values from a function. In this problem, referring to the image attached , we can see that the , we have some or other value of y for every x . That means that, it is defined for every real value in x. Hence the Domain will be Set of all real numbers . Where as the maximum value of y we have in this problem is 36. But y has no end towards negative side. Hence the possible outcomes are from negative infinity to 36. Hence range will be (-∞,36] , where 36 is included.

Domain = R

Range = (-∞,36]

b) x and y intercepts represents the distance of points where a graph cuts the both axes from the center. Please refer to the image attached. Here we can see that for our function $y=-x^2+36$

X intercepts are = ±6 as they cut x axis on two points ( -6,0) and (6,0)

Y intercept = 36 as graph cuts y axis at one point (0,36)

c) The linear function will the equation of line passing through (-5,11) and (4,20)

The slope of line will be

$m=\frac{20-11}{4-(-5)}$

m=1

Hence the function has a positive slope here , and hence we can say that it is a positive function.

The equation will be

$\frac{y-20}{x-4}=1$

y-20=x-4

$y=x+16$

d)

now we are having two equations.

i) $y=-x^2+36$

ii) $y=x+16$

The solution of above two system of linear equation will be the points at which they intersect. Referring to the image attached we can see that they are intersecting at (-5,11) and (4,20) . Hence x=-5 & 4 are the solution to the system of equation we have now.

Question 3:

Please refer to the Image attached for the piece wise function

Here we have assumed two function with condition like given under

y=x+1 { x ≤ 3 }

y=4 {x > 3}

for values less than or equal to 3 , we have our function as y=x+1

let us graph this by joining two coordinates

x=0, y =1

x=1 , y= 2

x=3 , y=4

Hence we have two coordinates as (0,1) and (1,2)

we join them and limit it upto only (3,4)

because for values x>3 , the function will result us , y=4

which is a constant function , hence for every value of x . 3 , we will have 4 as output. Hence it will represent a straight line parallel to the x axis at a distance of 4 units from it.

7 0

3 years ago