Answer
a. $t_{C}$ = $20^{0}$C
b. $t_{C}$ = $-31^{0}$C
c. $t_{F}$ = $79^{0}$F
d. $t_{F}$ = $-114^{0}$F
Work Step by Step
a. We can convert temperature in Fahrenheit into Celsius, by applying the formula:
$t_{C}$ = $\frac{5^{0}C}{9^{0}F}$ x ($t_{F}$-$32^{0}$F)
We know $t_{F}$ = $68^{0}$F , so we can write:
$t_{C}$ = $\frac{5^{0}C}{9^{0}F}$ x ($68^{0}$F-$32^{0}$F)
$t_{C}$ = $\frac{5^{0}C}{9^{0}F}$ x$36^{0}$F
$t_{C}$ = $20.0^{0}$C
$t_{C}$ = $20^{0}$C
b. Apply the formula:
$t_{C}$ = $\frac{5^{0}C}{9^{0}F}$ x ($t_{F}$-$32^{0}$F)
We know $t_{F}$ = $-23^{0}$F , so we can write:
$t_{C}$ = $\frac{5^{0}C}{9^{0}F}$ x ($-23^{0}$F-$32^{0}$F)
$t_{C}$ = $\frac{5^{0}C}{9^{0}F}$ x($-55^{0}$)F
$t_{C}$ = $-30.56^{0}$C
$t_{C}$ = $-31^{0}$C
c. Apply the formula to convert Ceslius to Fahrenheit:
$t_{F}$ =($t_{C}$ x$\frac{9^{0}F}{5^{0}C}$) +$32^{0}$F
We know $t_{C}$ = $26^{0}$C, so :
$t_{F}$ =($26^{0}$C x$\frac{9^{0}F}{5^{0}C}$) +$32^{0}$F
$t_{F}$ = $46.8^{0}$F+$32^{0}$F
$t_{F}$ = $78.8^{0}$F
$t_{F}$ = $79^{0}$F
d. Apply the formula to convert Ceslius to Fahrenheit:
$t_{F}$ =($t_{C}$ x$\frac{9^{0}F}{5^{0}C}$) +$32^{0}$F
We know $t_{C}$ = $-81^{0}$C, so :
$t_{F}$ =($-81^{0}$C x$\frac{9^{0}F}{5^{0}C}$) +$32^{0}$F
$t_{F}$ = $-145.8^{0}$F+$32^{0}$F
$t_{F}$ = $-113.8^{0}$F
$t_{F}$ = $-114^{0}$F
