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3 years ago

Please Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

MrMuchimi3 years ago

3 0

The cross-section is a rectangle. Its height is the same as the height of the prism, while its length is not necessarily the same but certainly perpendicular to the height.

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Find the values of x and y

Igoryamba

17y-8=12+13y
4y=20
y=5

11x-4=117
11x=121
x=11

This is the answer, and I have steps, if have any questions, message me.

5 0

3 years ago

True or False? The circumcenter of a triangle is the center of the only circle that can be inscribed about it.

zlopas [31]

It is False, for apex.

8 0

3 years ago

PLEASE HELP!! within the interval notation of [-3,-2) the function below is…

kkurt [141]

Answer: increasing

Step-by-step explanation: The slope of the graph is positive.

7 0

3 years ago

Tom is the deli manager at a grocery store.He needs to schedule employees to staff the deli department at least 260 person-hours

Drupady [299]

I hope this is right but:
Let n=number of full time employees

260<=20+n*40
260-20<=N*40
240<=n*40
n>=240/40
n>=6 full time employees

which means he would have to hire at least 6 or more fully time employees

Though I'm not sure if the attached file is the graph you are looking for

4 0

3 years ago

The graph of the continuous function g, the derivative of the function f, is shown above. The function g is piecewise linear for

4vir4ik [10]

A) $g=f'$ is continuous, so $f$ is also continuous. This means if we were to integrate $g$ , the same constant of integration would apply across its entire domain. Over $0$ , we have $g(x)=2x$ . This means that

$f_{0$

For $f$ to be continuous, we need the limit as $x\to1^-$ to match $f(1)=3$ . This means we must have

$\displaystyle\lim_{x\to1}x^2+C=1+C=3\implies C=2$

Now, over $x$ , we have $g(x)=-3$ , so $f_{x$ , which means $f(-5)=17$ .

b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,

$\displaystyle\int_1^6g(x)=4+\int_3^62(x-4)^2\,\mathrm dx=4+6=10$

c) $f$ is increasing when $f'=g>0$ , and concave upward when $f''=g'>0$ , i.e. when $g$ is also increasing.

We have $g>0$ over the intervals $0$ and $x>4$ . We can additionally see that $g'>0$ only on $0$ and $x>4$ .

d) Inflection points occur when $f''=g'=0$ , and at such a point, to either side the sign of the second derivative $f''=g'$ changes. We see this happening at $x=4$ , for which $g'=0$ , and to the left of $x=4$ we have $g$ decreasing, then increasing along the other side.

We also have $g'=0$ along the interval $-1$ , but even if we were to allow an entire interval as a "site of inflection", we can see that $g'>0$ to either side, so concavity would not change.

5 0

3 years ago