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3 years ago

Please help, Describe the transformation of the graph of the parent function y = √x for the function y = √x + 7 + 5.

Mathematics

2 answers:

KengaRu [80]3 years ago

8 0

Answer:

C. Up

2nd part: the third on

3rd part: the second one

Setler79 [48]3 years ago

5 0

Answer:

C. Up.

Step-by-step explanation:

We have been given that the parent function $y=\sqrt{x}$ is transformed to the function $y=\sqrt{x+7}+5$ . We are asked to describe the given transformation.

Let us recall the transformation rules.

$f(x+a)=\text{Graph shifted to left by a units}$ ,

$f(x-a)=\text{Graph shifted to right by a units}$ ,

$f(x)+a=\text{Graph shifted upwards by a units}$ ,

$f(x)-a=\text{Graph shifted downwards by a units}$ .

Upon comparing our transformed function with the parent function, we can see that the graph is shifted to left by 7 units and upwards by 5 units.

Therefore the correct choice is option C.

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3 years ago

A tank contains 5,000 L of brine with 13 kg of dissolved salt. Pure water enters the tank at a rate of 50 L/min. The solution is

tresset_1 [31]

Answer:

a) $x(t) = 13*e^(^-^\frac{t}{100}^)$

b) 10.643 kg

Step-by-step explanation:

Solution:-

- We will first denote the amount of salt in the solution as x ( t ) at any time t.

- We are given that the Pure water enters the tank ( contains zero salt ).

- The volumetric rate of flow in and out of tank is V(flow) = 50 L / min

- The rate of change of salt in the tank at time ( t ) can be expressed as a ODE considering the ( inflow ) and ( outflow ) of salt from the tank.

- The ODE is mathematically expressed as:

$\frac{dx}{dt} =$ ( salt flow in ) - ( salt flow out )

- Since the fresh water ( with zero salt ) flows in then ( salt flow in ) = 0

- The concentration of salt within the tank changes with time ( t ). The amount of salt in the tank at time ( t ) is denoted by x ( t ).

- The volume of water in the tank remains constant ( steady state conditions ). I.e 10 L volume leaves and 10 L is added at every second; hence, the total volume of solution in tank remains 5,000 L.

- So any time ( t ) the concentration of salt in the 5,000 L is:

$conc = \frac{x(t)}{1000}\frac{kg}{L}$

- The amount of salt leaving the tank per unit time can be determined from:

salt flow-out = conc * V( flow-out )

salt flow-out = $\frac{x(t)}{5000}\frac{kg}{L}*\frac{50 L}{min}\\$

salt flow-out = $\frac{x(t)}{100}\frac{kg}{min}$

- The ODE becomes:

$\frac{dx}{dt} = 0 - \frac{x}{100}$

- Separate the variables and integrate both sides:

$\int {\frac{1}{x} } \, dx = -\int\limits^t_0 {\frac{1}{100} } \, dt + c\\\\Ln( x ) = -\frac{t}{100} + c\\\\x = C*e^(^-^\frac{t}{100}^)$

- We were given the initial conditions for the amount of salt in tank at time t = 0 as x ( 0 ) = 13 kg. Use the initial conditions to evaluate the constant of integration:

$13 = C*e^0 = C$