All categories

1 year ago

If cos43=p, determine.without the use of a calculator,the value of each of the following in terms of p

Mathematics

1 answer:

I am Lyosha [343]1 year ago

3 0

We need to calculate the value of cos(43) to determine the value of p

<h3>How to evaluate the expression?</h3>

The equation is given as:

cos43 = p

Rewrite properly as:

cos(43) = p

Rewrite the equation as:

p = cos(43)

The above means that we simply calculate the value of cos(43) to determine the value of p

You might be interested in

How do cnidaria feed,move, and reproduce

Nady [450]

Answer:

Cnidarians do not have true organs, however. Reproduction is by asexual budding (polyps) or sexual formation of gametes (medusae, some polyps). The result of sexual reproduction is a larva, which can swim on its own.

Step-by-step explanation:

8 0

3 years ago

In \triangle LMN,△LMN, \angle N \cong \angle M,∠N≅∠M, LM = 14LM=14 and MN = 8MN=8. Find the length of NL.NL.

Black_prince [1.1K]

Answer:

11.5

Step-by-step explanation:

We solve using :

Pythagoras Theorem

LM² = MN² + NL²

NL² = LM² - MN²

NL = √LM² - MN²

NL = √14² - 8²

NL = √(132)

NL = 11.489125293

NL = 11.5

Therefore, the length of NL = 11.5

7 0

2 years ago

Needing help with this!

Sonbull [250]

assuming the scale is by 1,

a) the slope is rise/run so it rises 3 units and moves 4 units right therefore the slope is 3/4

b) the y intercept is where the line crosses the y axis so by looking at the graph you can tell that the y int is 1

c) and the equation is made up of the slope and the y intercept: y= 3/4x + 1

6 0

3 years ago

A tank contains 60 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

MissTica

Answer:

(a) 60 kg; (b) 21.6 kg; (c) 0 kg/L

Step-by-step explanation:

(a) Initial amount of salt in tank

The tank initially contains 60 kg of salt.

(b) Amount of salt after 4.5 h

$\text{Let A = mass of salt after t min}\\\text{and }r_{i} = \text{rate of salt coming into tank}\\\text{and }r_{0} =\text{rate of salt going out of tank}$

(i) Set up an expression for the rate of change of salt concentration.

$\dfrac{\text{d}A}{\text{d}t} = r_{i} - r_{o}\\\\\text{The fresh water is entering with no salt, so}\\ r_{i} = 0\\r_{o} = \dfrac{\text{3 L}}{\text{1 min}} \times \dfrac {A\text{ kg}}{\text{1000 L}} =\dfrac{3A}{1000}\text{ kg/min}\\\\\dfrac{\text{d}A}{\text{d}t} = -0.003A \text{ kg/min}$

(ii) Integrate the expression

$\dfrac{\text{d}A}{\text{d}t} = -0.003A\\\\\dfrac{\text{d}A}{A} = -0.003\text{d}t\\\\\int \dfrac{\text{d}A}{A} = -\int 0.003\text{d}t\\\\\ln A = -0.003t + C$

(iii) Find the constant of integration

$\ln A = -0.003t + C\\\text{At t = 0, A = 60 kg/1000 L = 0.060 kg/L} \\\ln (0.060) = -0.003\times0 + C\\C = \ln(0.060)$

(iv) Solve for A as a function of time.

$\text{The integrated rate expression is}\\\ln A = -0.003t + \ln(0.060)\\\text{Solve for } A\\A = 0.060e^{-0.003t}$

(v) Calculate the amount of salt after 4.5 h

a. Convert hours to minutes

$\text{Time} = \text{4.5 h} \times \dfrac{\text{60 min}}{\text{1h}} = \text{270 min}$

b.Calculate the concentration

$A = 0.060e^{-0.003t} = 0.060e^{-0.003\times270} = 0.060e^{-0.81} = 0.060 \times 0.445 = \text{0.0267 kg/L}$

c. Calculate the volume

The tank has been filling at 6 L/min and draining at 3 L/min, so it is filling at a net rate of 3 L/min.

The volume added in 4.5 h is

$\text{Volume added} = \text{270 min} \times \dfrac{\text{3 L}}{\text{1 min}} = \text{810 L}$

Total volume in tank = 1000 L + 810 L = 1810 L

d. Calculate the mass of salt in the tank

$\text{Mass of salt in tank } = \text{1810 L} \times \dfrac{\text{0.0267 kg}}{\text{1 L}} = \textbf{21.6 kg}$

(c) Concentration at infinite time

$\text{As t $\longrightarrow \, -\infty,\, e^{-\infty} \longrightarrow \, 0$, so A $\longrightarrow \, 0$.}$

This makes sense, because the salt is continuously being flushed out by the fresh water coming in.

The graph below shows how the concentration of salt varies with time.

3 0

2 years ago

Write the summation to estimate the area under the curve y = 1 + x^2 from x = –1 to x = 2 using 3 rectangles and right endpoints

Minchanka [31]

Check the picture below.

8 0

3 years ago