Simplify the following expression: NEED HELP QUICK PLZ :(

A ship sails from harbour h on a bearing of 084 for 340 km until it reaches point P. It the sails on a bearing of 210 for 160km

sattari [20]

Answer:

QH = 227.8 km ≅ 228 km

Step-by-step explanation:

∵ The bearing from H to P is 084°

∵ The bearing from P to Q is 210°

∵ The distance from H to P = 340 km

∵ The distance from P to Q = 160 km

∴ The angle between 340 and 160 = 360 - 210 - (180 - 84) = 54°

( 180 - 84) ⇒ interior supplementary

By using cos Rule:

(QH)² = (PH)² + (PQ)² - 2(PH)(PQ)cos∠HPQ

(QH)² = 340² + 160² - 2(340)(160)cos(54) = 51904.965

∴ QH = 227.8 km ≅ 228 km

7 0

3 years ago

Read 2 more answers

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

4 0

3 years ago

Combine like terms to create an equivalent expression. -3.6-1.9t+1.2+5.1t−3.6−1.9t+1.2+5.1t

Andreyy89

Question

Combine like terms to create an equivalent expression.

-3.6-1.9t+1.2+5.1t

Answer:

3.2t - 2.4

Step-by-step explanation:

Given;

-3.6 - 1.9t + 1.2 + 5.1t

Combining like terms means bringing terms that have "t" together and separately, those that don't have "t" together. i.e

=> − 1.9t + 5.1t - 3.6 + 1.2

=> 3.2t - 2.4

Therefore, the equivalent expression is;

3.2t - 2.4

4 0

3 years ago

there is a canned food sale at the supermarket. a case of 24 cans of peas is prices at $19.68. at the same rate what is the pric

Nonamiya [84]

Answer:

8.2

Step-by-step explanation:

We can find the unit rate of this problem by dividing 18.68 by 24 to get .82. .82*10 is equal to 8.2.

5 0

1 year ago

A simple random sample was taken of 44 water bottles from a bottling plant’s warehouse. The dissolved oxygen content (in mg/L) w

Margarita [4]

Answer:

(a) Test statistic is -2.85 and p-value is 0.0022

(b) Reject the null hypothesis. The population mean of dissolved oxygen content is not equal to 10 mg/L

Step-by-step explanation:

H0: mu equals 10

Ha: mu not equals 10

The test is a two-tailed test because the alternate hypothesis is expressed using not equal to

(a) Test statistic (z) = (sample mean - population mean) ÷ (sd/√n) = (9.14 - 10) ÷ (2/√44) = -0.86 ÷ 0.302 = -2.85

Cumulative area of the test statistic = 0.9978

p-value = 2(1 - 0.9978) = 2(0.0022) = 0.0044

(b) The critical value using 0.02 significance level is 2.422. For a two-tailed test, the region of no rejection of the test statistic lies between -2.422 and 2.422.

Conclusion:

Reject the null hypothesis because the test statistic -2.85 falls outside the region bounded by the critical values -2.422 and 2.422.

The population mean of dissolved oxygen content is not equal to 10 mg/L

7 0

3 years ago