3 years ago

Can someone help me please?

Mathematics

2 answers:

FromTheMoon [43]3 years ago

3 0

Answer:

the first one and the last three

Step-by-step explanation:

that should be it.

12345 [234]3 years ago

3 0

A, e,d,f........................

You might be interested in

Which ordered pairs in the form (x, y) are solutions to the equation

inna [77]

Answer:

(-1, 7) and (-6, -14) are solutions

4 0

3 years ago

Read 2 more answers

8(1/2u+3/4)

netineya [11]

Answer:

4u+3

Step-by-step explanation: 8/1*1/2= 8/2=4u

8/1*3/4= 24/8=3

4 0

3 years ago

How do you solve this ? what do you do with the exponets ?

antiseptic1488 [7]

4/7 is the answer.

I cant reall explain becuase it will take forever xD

6 0

3 years ago

Read 2 more answers

Soire 4(x + 1) + 8 = 24. (I point) оа A 3.75 B -2 C 5 D 3

Olin [163]

Answer:

Step-by-step explanation:

4(x + 1) + 8 = 24

Subtract 8 from each side

4(x + 1) + 8-8 = 24-8

4(x + 1) = 16

Divide by 4

4(x+1)/4 = 16/4

x+1 = 4

Subtract 1

x+1-1 = 4-1

x=3

7 0

3 years ago

Consider the functions below. f(x, y, z) = x i − z j + y k r(t) = 4t i + 6t j − t2 k (a) evaluate the line integral c f · dr, wh

fredd [130]

With

$\vec r(t)=4t\,\vec\imath+6t\,\vec\jmath-t^2\,\vec k$

we have

$\mathrm d\vec r=(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt$

The vector field evaluated over this parameterization is

$\vec f(x,y,z)=\vec f(x(t),y(t),z(t))=4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k$

so the line integral is

$\displaystyle\int_{-1}^1(4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k)\cdot(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt$

$=\displaystyle\int_{-1}^1(16t+6t^2-12t^2)\,\mathrm dt=-4$

6 0

3 years ago