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4 years ago

6

Given Sine (t) = StartFraction 7 Over 25 EndFraction for StartFraction pi Over 2 EndFraction less-than t less-than pi, what is C

osine (t)?

1 answer:

lys-0071 [83]4 years ago

6 0

Answer:

-24/25

Step-by-step explanation:

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Solve: 12 dogs decreased by 25% step by step pls.

Annette [7]

Decreased by 25%
find 25% of 12 and decrease 12 by it

percent means parts out of 100
25%=25/100=1/4
'of' means multiply
25% of 12=1/4 times 12=3

decrease 12 by 3
12-3=9

answer is 9

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3 years ago

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Write a problem that can be solved with the equation r divided by 6 = 14

Sergeu [11.5K]

My answer to you is R / 6 = 14

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3 years ago

Hermione buys 9/10 of a pound of plums. The plums cost $1.30 per pound. How much does Hermione spend on plums? $0.90 $1.17 $1.30

qwelly [4]

<h3>Answer:</h3>

$\large\boxed{\$1.17}$

<h3>Step-by-step explanation:</h3>

In this question, we're trying to find how much Hermione spent on plums.

We know that:

Hermione bought 9/10 of a pound of plums
Plums cost $1.30 <em>per pound</em>

With the information above, we can solve the question.

Since we know that plums cost $1.30 per pound and Hermione bought 9/10 of a pound, we would multiply 1.30 by 0.90 in order to get the price Hermione bought the plums for. Hermione is pretty much paying 90% of the cost of a pound.

$1.30*0.90=1.17$

When you multiply, you should get 1.17

This means that Hermione spent $1.17 on 9/10 pound of plums.

<h3>I hope this helped you out.</h3><h3>Good luck on your academics.</h3><h3>Have a fantastic day!</h3>

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3 years ago

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Put the numbers in order from least to greatest.

bixtya [17]

So if u simplify them all it would be 5000000, 4000000, 500000, 6000000, 7500000 so it would be least is 500000, 4000000, 5000000, 6000000, and 75000000 being the greatest... I hope this helps uuuu! :)

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3 years ago

The population of a certain city has been increasing exponentially accordingto the functionP (t) = 800, 000e0.051where t is the

kogti [31]

$\begin{gathered} \text{ (a) }Given\text{ that p(t) = }800,000e^{0.05t} \\ \text{ In the year 2014, t=0} \\ \text{substitute t=0 into p(t) above} \\ p(0)\text{ = 800000}e^{0.05(0)} \\ p(0)\text{ = 800000}e^0 \\ p(0)\text{ = 800000}(1) \\ p(0)\text{ = 800000} \\ \text{The population at 2014 is 800,000} \end{gathered}$

(b) To determine the year the population will reach 1,200,000, substitute p(t) = 1,200,000 into p(t) above and calculate the value of t

$\begin{gathered} \text{ p(t) = 1,200,000} \\ 1,200,000=800,000e^{0.05t} \\ \text{divide both sides by 800,000} \\ \frac{1200000}{800000}\text{ =}\frac{\text{ 800000}e^{0.05t}}{800000} \\ \\ 1.5=e^{0.05t} \\ \text{take }\ln \text{ of both sides} \\ \ln 1.5\text{ = }\ln e^{0.05t} \\ \ln 1.5=0.05t\ln e \\ \ln 1.5=0.05t(1) \\ \ln 1.5\text{ = 0.05t} \\ \text{divide both sides by 0.05} \\ \frac{\ln 1.5}{0.05}\text{ = t} \\ t=8 \end{gathered}$

The year when t= 8, given that t=0 in 2014 is the year 2022

(c) the rate of change of the population in the year 2019

Firstly, differentiate p(t) with respect to t

$\begin{gathered} p(t)\text{ = }800,000e^{0.05t} \\ \frac{\text{ dp}}{\text{ dt}}\text{ = 0.05(800,000)}e^{0.05t} \\ \frac{dp}{\mathrm{d}t}\text{ = 40,000}e^{0.05t} \\ \text{ This means that the rate of change of the function at any time t is 40,000}e^{0.05t} \\ \end{gathered}$

In the year 2019, t = 5

Substitute t=5 into the rate of change above

$\begin{gathered} \text{ at t= 5, } \\ \text{rate of change = 40,000}e^{0.05(5)} \\ =\text{ 40,000}e^{0.25} \\ =40,000(1.284025) \\ =51361 \end{gathered}$

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2 years ago