What's the correct value for the expression 3 x |-4 + 2| - 3 ?

1 answer:

meriva3 years ago

5 0

In PEMDAS, we start with parenthesis (absolute values count as this), which means that |-4+2|= |-2|=2, resulting in 3*2-3=6-3=3

You might be interested in

What is the slope, m, of the line through (-7, -2) and (-6, 7)?

Nataliya [291]

Answer:

Slope = 9

Step-by-step explanation:

(-7 , -2) ; x₁ = -7 and y₁ = -2

(-6 , 7) ; x₂ = -6 and y₂ = 7

Slope = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$=\frac{7-[-2]}{-6-[-7]}\\\\=\frac{7+2}{-6+7}\\\\=\frac{9}{1}\\\\= 9$

4 0

3 years ago

Rosa wants to use $20 to buy games. The

kompoz [17]

Answer:

The answer is B : 6

Step-by-step explanation:

2.50k means 2.50 multiplied by a number and the inequality means 20 is less than 2.50k + 5.50. 2.50 multiplied by 6 equals 15 and 15 plus 5.50 is 20.50 which is greater than 20.

6 0

3 years ago

Find the flux of the following vector fields across the given surface with the specified orientation. You may use either an expl

BlackZzzverrR [31]

The tetrahedron passes through the intercepts (5, 0, 0), (0, 2, 0), and (0, 0, 10). Parameterize the surface (call it $\Sigma$ ) by

$\vec r(u,v)=(1-v)\langle5,0,0\rangle+v\left((1-u)\langle0,2,0\rangle+u\langle0,0,10\rangle\right)$

$\vec r(u,v)=\langle5(1-v),2(1-u)v,10uv\rangle$

with $0\le u\le1$ and $0\le v\le1$ . Take the normal vector to $\Sigma$ to be

$\vec r_v\times\vec r_u=\langle20v,50v,10v\rangle$

Then the flux of $\vec F(x,y,z)=\langle x,y,z\rangle$ across $\Sigma$ is

$\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1\langle5(1-v),2(1-u)v,10uv\rangle\cdot\langle20v,50v,10v\rangle\,\mathrm du\,\mathrm dv$