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svet-max [94.6K]

3 years ago

15

The elevation E, in meters, above sea level at which the boiling point of a certain liquid is t degrees Celsius is given by the

function shown below. At what elevation is the boiling point 119.5? 120? E(t)= 1200(120-t)+520(120-t)^2 At what elevation is the boiling point 119.5?
E(119.5)= (the answer) meters

1 answer:

Anna007 [38]3 years ago

5 0

Answer:

$E(119.5)= 730m$

Step-by-step explanation:

Given

$E(t)= 1200(120-t)+520(120-t)^2$

Required

Determine E(119.5)

$E(t)= 1200(120-t)+520(120-t)^2$

Simply substitute 119.5 for t

$E(119.5)= 1200(120-119.5)+520(120-119.5)^2$

Evaluate the expressions in bracket

$E(119.5)= 1200(0.5)+520(0.5)^2$

Solve 0.5²

$E(119.5)= 1200 * 0.5+520*0.25$

$E(119.5)= 600+130$

$E(119.5)= 730$

Hence;

$E(119.5)= 730m$

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Prove the divisibility:<br><br>45^45·15^15 by 75^30

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Substituting these values in our division problem we will get,

$\frac{9^{45}*5^{45}*3^{15}*5^{15}}{15^{30}*5^{30}}$

Using power rule of exponents $a^n*a^m=a^{n+m}$ we will get,

$\frac{9^{45}*5^{(45+15)}*3^{15}}{15^{30}*5^{30}}$

$\frac{9^{45}*5^{60}*3^{15}}{15^{30}*5^{30}}$

Using product rule of exponents $(a*b)^n=a^n*b^n$ we will get,

$\frac{(3*3)^{45}*5^{60}*3^{15}}{(3*5)^{30}*5^{30}}$

$\frac{3^{45}*3^{45}*5^{60}*3^{15}}{3^{30}*5^{30}*5^{30}}$

Using power rule of exponents $a^n*a^m=a^{n+m}$ we will get,

$\frac{3^{(45+45+15)}*5^{60}}{3^{30}*5^{(30+30)}}$

$\frac{3^{105}*5^{60}}{3^{30}*5^{60}}$

$\frac{3^{105}}{3^{30}}$

Using quotient rule of exponent $\frac{a^m}{a^n}=a^{m-n}$ we will get,

$\frac{3^{105}}{3^{30}}=3^{105-30}$

$3^{105-30}=3^{75}$

Therefore, our resulting quotient will be $3^{75}$ .

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