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3 years ago

10

On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snoho

mish River below. The swimmers stepped off the bridge, and I estimated that they hit the water 1.5 s later. (a) How high was the bridge

1 answer:

tamaranim1 [39]3 years ago

4 0

Answer: the bridge is 11.025 m high

Step-by-step explanation:

Given that;

Time taken to hit the water t = 1.5 sec

height of bridge = ?

lets take a look at the equation of motion;

y(t) = y₀ + (1/2)at²

with initial velocity zero and we know that acceleration due to gravidity is 9.8m/s

we substitute

y(t) = (1/2)gt² = (1/2) × 9.8 × (1.5)²

y(t) = (1/2)gt² = (1/2) × 9.8 × (1.5)²

y(1.5) = 11.025 m

Therefore, the bridge is 11.025 m high

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3 years ago

If logb2=x and logb3=y, evaluate the following in terms of x and y:

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$log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x$

<em><u>Solution:</u></em>

Given that,

$log_b2 = x\\\\log_b3 = y --------(i)$

<em><u>Use the following log rules</u></em>

Rule 1: $log_b(ac) = log_ba + log_bc$

Rule 2: $log_b\frac{a}{c} = log_ba - log_bc$

Rule 3: $log_ba^c = clog_ba$

$a) log_b{162}$

Break 162 down to primes:

$162 = 2^1 \times 3^4$

$log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y$

Thus we get,

$log_b162 = x + 4y$

Next

$b) log_b 324$

Break 324 down to primes:

$324 = 2^2 \times 3^4$

$log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y$

Next

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By rule 2

$log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} = 3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} = 3x - 2y$

Next

$d) \frac{log_b27}{log_b16}$

By rule 2

$\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} = 3y - 4x$

Thus the given are evaluated in terms of x and y

3 0

3 years ago

JulijaS [17]

Answer: $8865=6000(1+\frac{5\times t}{100})$

Step-by-step explanation:

Given

Liam deposited $P=\$\ 6000$

Rate of interest is $R=5\ \%$

If the amount after t years is $A=\$\ 8865$

Simple interest is given by

$S.I.=\dfrac{P\times R\times t}{100}$

And amount is

$A=P+S.I.$

$A=P(1+\frac{R\times t}{100})$

Substituting values we get

$8865=6000(1+\frac{5\times t}{100})$

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$5t=47.75$

$t=\frac{47.75}{5}=9.55\ years$

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