Please help I don't understand

Mathematics

1 answer:

Bond [772]3 years ago

3 0

Answer:

mx+b

Step-by-step explanation:

You might be interested in

Calculus: Help ASAP

Serggg [28]

$\bf \displaystyle \int\limits_{-1}^{0}~(4x^6+2x)^3(12x^5+1)\cdot dx\\\\
-------------------------------\\\\
u=4x^6+2x\implies \cfrac{du}{dx}=24x^5+2\implies \cfrac{du}{2(12x^5+1)}=dx\\\\
-------------------------------\\\\
\displaystyle \int\limits_{-1}^{0}~u^3\underline{(12x^5+1)}\cdot\cfrac{du}{2\underline{(12x^5+1)}}\implies \cfrac{1}{2}\int\limits_{-1}^{0}~u^3\cdot du\\\\
-------------------------------\\\\$

$\bf \textit{now, we'll change the bounds, using u(x)}
\\\\\\
u(-1)=4(-1)^6+2(-1)\implies u(-1)=2
\\\\\\
u(0)=4(0)^6+2()\implies u(0)=0\\\\
-------------------------------\\\\
\displaystyle \cfrac{1}{2}\int\limits_{2}^{0}~u^3\cdot du\implies \left. \cfrac{1}{2}\cdot \cfrac{u^4}{4} \right]_{2}^{0}\implies \left. \cfrac{u^4}{8} \right]_{2}^{0}\implies [0]-[2]\implies -2$

5 0

3 years ago

the diameter of a piston for an automobile is 3 11/16in with tolerence of 1/64in. find the upper and lower limits of the pistons

padilas [110]

Upper Tolerance

Remark

The 11/16 is the only thing that will be affected. The three won't go up or down when we add 1/64 so we should just work with the 11/16. We need only add 11/16 and 1/64 together to see what the upper range is. Later on we can add 3 into the mix.

Solution

<u>Upper Limit</u>

$\dfrac{11}{16} + \dfrac{1}{64}$