85% of z is 106,250. What is z?

Answer:

-8y +8v +24

Step-by-step explanation:

-8(y-v-3)

Multiply each term inside the parentheses by -8

-8y -v*-8 -3*-8

-8y +8v +24

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Hey!!!

solution,

-8(y-v-3)

= -8y+8v+24

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Here,

You have to remember these things:

Hope it helps.

Good luck on your assignment

Answer:

Step-by-step explanation:

[tex]\int\limits^{\infty}_0 {A^2_1} (e^{-r/a})r^2dr= {A^2_1}\int\limits^{\infty}_0r^2(e^{-r/a})^2\, dr)[/tex]

[tex]=A_1^2\int\limits^{\infty}_0 r^2e^{-2r/a}\ dr[/tex]

[tex]=A_1^2[\frac{r^2e^{2r/a}}{-2/a} |_0^{\infty}-\int\limits^{\infty}_0 2r\frac{e^{-2r/a}}{-2/a} \ dr][/tex]

[tex]=A^2_1[0+\int\limits^{\infty}_0 a\ r\ e^{-2r/a}\ dr][/tex]

[tex]=A^2_1[\frac{a \ r \ e^{-2r/a}}{-2/a} |^{\infty}_0-\int\limits^{\infty}_0 \frac{a \ e^{-2r/a}}{-2/a} \ dr][/tex]

[tex]=A_0^2[0-0+\int\limits^{\infty}_0 \frac{a^2}{2} e^{-2r/a}\ dr\\\\=A_1^2\frac{a^2}{2} \int\limits^{\infty}_0 e^{-2r/a}\ dr\\\\=A_1^2\frac{a^2}{2} [\frac{e^{-2r/a}}{-2/a} ]^{\infty}_0[/tex]

[tex]=\frac{A_1^2a^2}{2} -\frac{a}{2} [ \lim_{r \to \infty} [e^{-2r/a} -e^0]\\\\=\frac{A_1^2a^2}{2} -(\frac{a}{2}) (0-1)[/tex]

[tex]=\frac{A_1^2a^3}{4}[/tex]

[tex]\therefore A_1^2\int\limits^{\infty}_0 r^2(e^{-r/a}) \ dr =\frac{A_1^2a^3}{4}[/tex]

Find the unique positive value of A1

[tex]=4\pi (\frac{A_1^2a^3}{4} )\\\\=A_1^2a^3\pi\\\\A_1^2=\frac{1}{a^3\pi} \\\\A_1=\sqrt{\frac{1}{a^3\pi} }[/tex]

Answer:

Plan A would be the least expensive

Step-by-step explanation:

Plan A= $0.45x100= 45, 45+40=$85

Plan B= $0.35x100= 35, 35+70= %105

(Each plan is for 100 minutes)