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3 years ago

1 1/4×-1 1/3 is equals to?

Mathematics

1 answer:

slavikrds [6]3 years ago

3 0

Answer: $-\frac{5}{12}$

Steps:

$1\frac{1}{4}\times \left(-1\right)\frac{1}{3}$

Convert mixed numbers to improper fractions: $1\frac{1}{4}$ $=\frac{5}{4}$

$=\frac{5}{4}\times \left(-1\right)\frac{1}{3}$

$\mathrm{Apply\:the\:rule}:\quad \:a\left(-b\right)=-ab$

$\frac{5}{4}\left(-1\right)=-\frac{5}{4}\times \:1$

$=-\frac{5}{4}\times \:1\times \frac{1}{3}$

$\mathrm{Apply\:rule}\:a\times 1=a$

$=-\frac{5}{4}\times \frac{1}{3}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{b}\times \frac{c}{d}=\frac{a\:\times \:c}{b\:\times \:d}$

$\frac{5}{4}\times \frac{1}{3}=\frac{5\times \:1}{4\times \:3}$

$=-\frac{5\times \:1}{4\times \:3}$

$\mathrm{Multiply\:the\:numbers:}\:5\times \:1=5$

$=-\frac{5}{4\times \:3}$

$\mathrm{Multiply\:the\:numbers:}\:4\times \:3=12$

$=-\frac{5}{12}$

Hope This Helps!

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2 years ago

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

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i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
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We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

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3 years ago

1 1/4×-1 1/3 is equals to?​

1 1/4×-1 1/3 is equals to?