All categories

2 years ago

3-4(12 divided by 6)

Mathematics

2 answers:

pashok25 [27]2 years ago

8 0

Answer:

Step-by-step explanation:

Do 12/6 first and you get 2 and then you do 3-4 and you get 1

Alex2 years ago

5 0

Answer:

The answer is -2

Step-by-step explanation:

First we divide 12 by 6.

12/6 = 2

Next we subtract 3 by 4.

3 - 4 = -1

Now we solve.

-1 * 2 = -2

So, the answer is -2.

Hope this helps! :)

You might be interested in

Find f(4) if f(x)=−3x2+11x+37

amid [387]

Hi there!

$\large\boxed{f(4) = 33}$

f(x) = -3x² + 11x + 37

Substitute in 4 for x to find f(4):

f(4) = -3(4)² + 11(4) + 37

f(4) = -3(16) + 44 + 37

f(4) = -48 + 44 + 37

f(4) = 33

6 0

2 years ago

Read 2 more answers

What's an expression that can be used to multiply 6 x 198 mentally

disa [49]

Sense it is metal math you do 6 x 200
<span />

3 0

3 years ago

Read 2 more answers

Does the order pair, (2,3) satisfy the following function y=-3x-2 Yes,No

grin007 [14]

Answer:

Step-by-step explanation:

The function y=-3x-2 does not go through the ordered pair (2,3).

3 0

2 years ago

Answers please! will make you brainliest!

LenKa [72]

11. 8/7
14. 13/10
17. 11/14

6 0

3 years ago

Read 2 more answers

Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).

lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted $C$ , which we can parameterize by the vector-valued function,

$\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)$

for $0\le t\le1$ , which has differential

$\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

Then with $x(t)=y(t)=z(t)=1-t$ , we have

$\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r$

$=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt$

Complete the square in the quadratic term of the integrand: $t^2-2t+2=(t-1)^2+1=(1-t)^2+1$ , then in the integral we substitute $u=1-t$ :

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt$

$\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du$

Make another substitution of $v=u^2$ :

$\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv$

Integrate by parts, taking

$r=v+1\implies\mathrm dr=\mathrm dv$

$\mathrm ds=e^v\,\mathrm dv\implies s=e^v$

$\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv$

$\displaystyle=-(2e-1)+(e-1)=-e$

So, we have by the fundamental theorem of calculus that

$\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)$

$\implies-e=f(1,1,1)-2$

$\implies f(1,1,1)=2-e$

3 0

3 years ago