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Alexandra [31]
4 years ago
6

50PTS!! does anyone know how you turn in a section of your slides my teacher wants us to turn in only a portion of our google sl

ides because we work on them each week(she says it gets confusing) if you know how to do this please let me know and will mark brainliest :). ​
Computers and Technology
1 answer:
HACTEHA [7]4 years ago
5 0

Answer:

The best way to turn in one part of google slides, since you can't split the slide when turning assignments on classroom, is to copy and paste the part you want to turn in into another slide and submit that portion. I've done for many projects before and if you name the different slides (Ex: Assignment Portion One 5/6/20) so it won't get confusing with the multiple slides.

This is the best way I can think of, I hope I helped, and please correct me if there is a better way!

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All of the following are career pathways in the architecture and construction career cluster except
Dennis_Churaev [7]

Below, I believe are the multiple choices attached to this question

A. Power, Structural, and Technical Systems.

B. Construction.

C. Design/ Pre-Construction.

D. Maintenance/ Operations

The answer is A: Power, Structural, and Technical Systems.

The career pathways in the Architecture and Construction deal with all aspects of designing, planning, and maintaining all kinds of structures we live or work in. It is organized into 3 career pathways; Construction, Design/ PreConstruction, and Maintenance/ Operations. This field also covers the servicing of equipment such as plumbing, electrical wiring, escalators, and elevators.

8 0
3 years ago
Y’all think Super Drags is good?
Kryger [21]

Yeah, I'm into it. It does show a lot of stereotypical views on drag queens, and it goes a little over the top, but honestly? The LGBT community has spent so long acting like the general population, and we're expected to be a sort of cookie cutter outline of the ideal person in order to fit in. We're not really allowed to be silly and have fun, otherwise we just get labeled as a stereotype, which sucks. When you're queer, you get labeled as that before anything else: your interests are seen as a byproduct of your queerness, not as an interest. So Super Drags, is actually a nice sort of change of pace. It's silly, it shows that queer people are human, and it sorta shows that "Yass bih" look on life, which is hilarious imo. Plus hey, Brazilian LGBT show that doesn't spout homophobic propaganda and supports diversity within all aspects of life? I'll support that.

TLDR; There aren't many silly shows out there that have an LGBT cast. Like, it's always supposed to be grim and sad, and all about heartbreak and coming out, yadda yadda yadda. So, it's cool that we've finally got something lighthearted.

3 0
4 years ago
Determine and prove whether an argument in English is valid or invalid. About Prove whether each argument is valid or invalid. F
yawa3891 [41]

Answer:

Each understudy on the respect roll got an A.  

No understudy who got a confinement got an A.  

No understudy who got a confinement is on the respect roll.  

No understudy who got an A missed class.  

No understudy who got a confinement got an A.  

No understudy who got a confinement missed class  

Explanation:

M(x): x missed class  

An (x): x got an A.  

D(x): x got a confinement.  

¬∃x (A(x) ∧ M(x))  

¬∃x (D(x) ∧ A(x))  

∴ ¬∃x (D(x) ∧ M(x))  

The conflict isn't considerable. Consider a class that includes a lone understudy named Frank. If M(Frank) = D(Frank) = T and A(Frank) = F, by then the hypotheses are overall evident and the end is counterfeit. Toward the day's end, Frank got a control, missed class, and didn't get an A.  

Each understudy who missed class got a confinement.  

Penelope is an understudy in the class.  

Penelope got a confinement.  

Penelope missed class.  

M(x): x missed class  

S(x): x is an understudy in the class.  

D(x): x got a confinement.  

Each understudy who missed class got a confinement.  

Penelope is an understudy in the class.  

Penelope didn't miss class.  

Penelope didn't get imprisonment.  

M(x): x missed class  

S(x): x is an understudy in the class.  

D(x): x got imprisonment.  

Each understudy who missed class or got imprisonment didn't get an A.  

Penelope is an understudy in the class.  

Penelope got an A.  

Penelope didn't get repression.  

M(x): x missed class  

S(x): x is an understudy in the class.  

D(x): x got a repression.  

An (ax): x got an A.  

H(x): x is on the regard roll  

An (x): x got an A.  

D(x): x got a repression.  

∀x (H(x) → A(x)) a  

¬∃x (D(x) ∧ A(x))  

∴ ¬∃x (D(x) ∧ H(x))  

Real.  

1. ∀x (H(x) → A(x)) Hypothesis  

2. c is a self-self-assured element Element definition  

3. H(c) → A(c) Universal dispatch, 1, 2  

4. ¬∃x (D(x) ∧ A(x)) Hypothesis  

5. ∀x ¬(D(x) ∧ A(x)) De Morgan's law, 4  

6. ¬(D(c) ∧ A(c)) Universal dispatch, 2, 5  

7. ¬D(c) ∨ ¬A(c) De Morgan's law, 6  

8. ¬A(c) ∨ ¬D(c) Commutative law, 7  

9. ¬H(c) ∨ A(c) Conditional character, 3  

10. A(c) ∨ ¬H(c) Commutative law, 9  

11. ¬D(c) ∨ ¬H(c) Resolution, 8, 10  

12. ¬(D(c) ∧ H(c)) De Morgan's law, 11  

13. ∀x ¬(D(x) ∧ H(x)) Universal hypothesis, 2, 12  

14. ¬∃x (D(x) ∧ H(x)) De Morgan's law, 13  

4 0
3 years ago
Write a program that reads in an integer value for n and then sums the integers from n to 2 * n if n is nonnegative, or from 2 *
devlian [24]

Answer:

Following are the program in c++

First code when using for loop

#include<iostream> // header file  

using namespace std; // namespace

int main() // main method

{

int n, i, sum1 = 0; // variable declaration

cout<<"Enter the number: "; // taking the value of n  

cin>>n;

if(n > 0) // check if n is nonnegative

{

for(i=n; i<=(2*n); i++)

{

sum1+= i;

}

}

else //  check if n  is negative

{

for(i=(2*n); i<=n; i++)

{

sum1+= i;

}

}

cout<<"The sum is:"<<sum1;

return 0;

Output

Enter the number: 1

The sum is:3

second code when using while loop

#include<iostream> // header file  

using namespace std; // namespace

int main() // main method

{

int n, i, sum1 = 0; // variable declaration

cout<<"Enter the number: "; // taking the value of n  

cin>>n;

if(n > 0) // check if n is nonnegative

{

int i=n;  

while(i<=(2*n))

{

sum1+= i;

i++;

}

}

else //  check if n  is negative

{

int i=n;  

while(i<=(2*n))

{

sum1+= i;

i++;

}

}

cout<<"The sum is:"<<sum1;

return 0;

}

Output

Enter the number: 1

The sum is:3

Explanation:

Here we making 2 program by using different loops but using same logic

Taking a user input in "n" variable.

if n is nonnegative then we iterate the loops  n to 2 * n and storing the sum in "sum1" variable.

Otherwise iterate a loop  2 * n to n and storing the sum in "sum1" variable.  Finally display  sum1 variable .

4 0
3 years ago
Some personal computer manufacturers provide a hard disk configuration that connects multiple smaller disks into a single unit t
pav-90 [236]
In software: Logical Volume. In hardware it's usually called RAID (Redundant Array of Inexpensive Devices).
6 0
4 years ago
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