All categories

3 years ago

What is the value of -2xy if x = -1 and y= 6? A. 12 B. -12 C. 26 D. 8

Mathematics

1 answer:

almond37 [142]3 years ago

6 0

Answer:

A. 12

Step-by-step explanation:

-2*-1=2*6=12

plz mark brainliest

You might be interested in

for a field trip 4 students rode in cars and the rest filled nine buses. how many students were in each bus if 472 students were

Lemur [1.5K]

The answer is 52

x - the number of students in a bus

4 + 9x = 472
9x = 472 - 4
9x = 468
x = 468 / 9
x = 52

4 0

2 years ago

Read 2 more answers

2. Given square CANE with diagonals

svlad2 [7]

What’s something goes up but never comes down?

6 0

3 years ago

A group of 20 friends were going on a picnic 7 friends were playing basketball and 5 friends were having snacks if 2 friends wer

dlinn [17]

The problem statement tells you "7 friends were playing basketball".

_____
It appears 5 friends were only playing basketball, 2 friends were playing basketball and having snacks, and 3 friends were only having snacks. Apparently, 10 friends were doing neither of those activities.

4 0

3 years ago

Please help ASAP! Only one question!!

levacccp [35]

It is negative because the plot is going down not up

5 0

3 years ago

Leno4ka [110]

Answer:

$\frac{s^2-25}{(s^2+25)^2}$

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given: $\mathcal{L}[t \cos 5t]=(-1)F'(s)$ with $F(s)=\mathcal{L}[\cos 5t]$ .

Now, $F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt$ . Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that $F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt$ .

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

$F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s)$ .

Solving for F(s) on the last equation, $F(s)=\frac{s}{s^2+25}$ , then the Laplace transform we were searching is $-F'(s)=\frac{s^2-25}{(s^2+25)^2}$

3 0

2 years ago