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4 years ago

Guys I really need help with these too thanks u before!!!!

Mathematics

1 answer:

Natali5045456 [20]4 years ago

5 0

Hey!!

here is your answer >>>

1) 3x - 10 > 11

let's take this as 3x - 10 = 11

3x = 21

x = 7
So, The numbers of X are always more than 7

2) 9x + 2 < 5

9x + 2 = 5

9x = 3

x = 1÷2

The numbers of X are always less than 1 ÷ 2

3) 2x + 8 < 6

2x + 8 = 6

2x = 2

x = 1

The numbers of X are less than 1

4) 4x - 3 < 12

4x - 3 = 12

4x = 15

x = 15 ÷ 4

The numbers of X are less than 15 ÷ 4

5) 2 ( 9 + x ) > 20

2 ( 9 + x ) = 20

9 + x = 10

x = 1

The numbers of X are more than 1

We can solve the rest using this

Hope my answer helps!

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Please find attached to this answer a well labelled diagram of the triangle

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Approximately 176.6 feet

Step-by-step explanation:

From the question, we are to find the height at which the baseball fan was sitting.

From the attached diagram the height at which the fan was sitting is equivalent to the opposite side of the triangle.

Hence , the appropriate trigonometric function to use to solve this question is the sine function.

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Opposite is unknown so we refer to it as Y

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3 years ago

Cristina uses a ruler to measure the length of her math textbook. She says that the book is 4/10 meters long. Is her measurement

a_sh-v [17]

Answer:

2/5

Step-by-step explanation:

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3 years ago

How many pounds of iron are produced from 4,200 pounds of iron ore if the ratio of iron ore is 5/21

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4 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.

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3 years ago