All categories

2 years ago

[1 -4 3 5]+[-2 6 -2 4]

Mathematics

1 answer:

Rasek [7]2 years ago

4 0

Step-by-step explanation:

[1-2 -4+6 3-2 5+4]

[-1 2 1 9]

You might be interested in

suppose you are in your last semester of college and have a 2.75 GPA after 105 credit hours if you are taking 15 credit hours in

Valentin [98]

Answer:

The overall CGPA would be 2.90 so it is not possible for hum to secure a CGPA of 3 for graduation.

Step-by-step explanation:

Given,

CGPA = 2.75

Credit hours = 105

Last semester GPA = 4

Last semester credit hours = 15

CGPA = $\frac{sum of (Credit hours of each subject * GPA of each subject)}{Total credit hours}$ credit hours * gpa

in our case, it would be:

CGPA = $\frac{(2.75*105)+(4*15}{105+15}$

=> $\frac{(288.75)+(60)}{120}$

=> $\frac{348.75}{120}$

=> 2.90

The overall CGPA would be 2.90 so it is not possible for hum to secure a CGPA of 3 for graduation.

7 0

4 years ago

Please due in 15 minutes please please!!!!!!

liraira [26]

8 girls 12 boys = 20 total
12/20= 3/5 = .6 *100 =
60% is the first answer

34 trees (3 types) +? Scotch pines
= 50 total
50-34= 16/50*2 = 32/100=
32% scotch pines is the second answer

20 grapes 1 pear
13/20*5 = 65/100 =
65% grapes third answer

Found 20 | did not find 5 | 25 total
5/25*4= 20/100=
20% is the fourth answer

4 0

3 years ago

PLEASE ANSWER CORRECTLY:

KATRIN_1 [288]

Answer:

%45

Step-by-step explanation:

22 - 40

-18

=-45%

= 45% error

3 0

3 years ago

Jackson wants to tile the floor in his bathroom with square tiles that are 1/2 foot long. Will he use more or fewer tiles if he

andreev551 [17]

He will use more tiles because 1/3 is smaller than 1/2 and the smaller the tiles the more you need to fill the place up.

3 0

3 years ago

In a free-fall experiment, an object is dropped from a height of h = 400 feet. A camera on the ground 500 ft from the point of i

PilotLPTM [1.2K]

Hmmm the object, is at rest, when dropped, so it has a velocity of 0 ft/s

the only force acting on the object, is gravity, using feet will then be -32ft/s²,

was wondering myself on -32 or 32.. but anyhow... we'll settle for the negative value, since it seems to be just a bit of convention issues

so, we'll do the integral to get v(t) then

$\bf \displaystyle \int -32\cdot dt\implies -32t+C
\\\\\\
\textit{object moves from \underline{rest}, so velocity is 0 at 0secs}
\\\\\\
-32(0)+C=0\implies C=0\implies \boxed{v(t)=-32t}
\\\\\\
\textit{now to get the positional s(t)}
\\\\\\
\displaystyle \int -32t\cdot dt\implies -16t^2+C
\\\\\\
\textit{the initial \underline{position} was 400ft away at 0secs}
\\\\\\
-16(0)^2+C=400\implies C=400\implies \boxed{s(t)=-16t^2+400}$

when will it reach the ground level? let's set s(t) = 0

$\bf s(t)=-16t^2+400\implies 0=-16t^2+400\implies \cfrac{-400}{-16}=t^2
\\\\\\
25=t^2\implies \boxed{5=t}$

part B) check the picture below

5 0

3 years ago