Translate to an algebraic equation<br> 5 less than one-half x is fifteen.

Ok so, based on the graph lets say that x = seconds and y = depth of dolphin. the interception in both points is when they are at 0 i.e when x is equals 0 and y equals 0. So At 0 seconds, the dolphin is 42 feet below the surface. So we say that x = 0, y = -42 and then the y intercepts = -42 so the point of interception is(0 (seconds),-42(depth of the dolphin)) When the clock says it's 14 seconds, the dolphin is even with the surface (0 below the surface this time). So we say that x = 14, y = 0. In this case The x intercept = 14 (14,0). But we need to calculate The slope with the formula= (y2 - y1)/(x2 - x1) = 42/14 = 3. Therefore, the formula for this line is y = 3x - 42.

3 0

4 years ago

Browning Labs is testing a new growth inhibitor for a certain type of bacteria. The bacteria naturally grows exponentially each

makvit [3.9K]

Hey! I just answered this on plato. the answer is that it includes negative factors, which makes not all solutions viable.

3 0

3 years ago

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If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,

julsineya [31]

<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

$\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)$

$(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166$

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0

4 years ago

Translate to an algebraic equation 5 less than one-half x is fifteen.