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2 years ago

What is the range of the function y= √x+5

Mathematics

2 answers:

Fofino [41]2 years ago

5 0

Answer: {f(x)∈R∣f(x)≥0}

Step-by-step explanation: The square root function never produces a negative result. Therefore, for the function f(x)=√x+5 , the domain is {x∈R∣x≥−5} and the range is {f(x)∈R∣f(x)≥0} .

Tema [17]2 years ago

3 0

Answer:

Y>0

Step-by-step explanation:

To determine the range of this function, we must first evaluate the domain. The square root function is a nice, neat function as long as the radicand isn’t negative. In this function, the radicand becomes negative after x gets smaller than -5, so the domain of this function is [-5, infinity).

Now that we know the domain, we can calculate the range. Beginning with the left boundary, we can substitute -5 into the function to see what y equals at this x-value. At -5, y equals 0, so the minimum value for the range is 0; with the right boundary, substituting infinity yields infinity, so the range is any number greater than 0.

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2 divided by 5/12= what

AlekseyPX

Answer:

4.8

Step-by-step explanation:

2 ÷ 5/12 = 4.8

Make sure to use a calculator or solve on a piece of paper. :)

6 0

3 years ago

Riley has collected surveys for her final thesis project. One one question, 46.1% of respondents said that conditions were impro

ira [324]

Answer:

The maximum value of the confidence interval for this set of survey results is 51.73%.

Step-by-step explanation:

A confidence interval has two bounds, a lower bound and an upper bound.

These bounds depend on the sample proportion and on the margin of error.

The lower bound is the sample proportion subtracted by the margin of error.

The upper bound is the margin of error added to the sample proportion.

In this question:

Sample proportion: 46.1%

Margin of error: 5.63%.

Maximum value is the upper bound:

46.1+5.63 = 51.73

The maximum value of the confidence interval for this set of survey results is 51.73%.

5 0

3 years ago

Solve for a:<br><br> 28000/1+a = 0

nika2105 [10]

1207838929299236636236+2892

5 0

3 years ago

The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature

Karolina [17]

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

$T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$

Where

$(t) = \ time$

$T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)$

$T_{s} = Surrounding \ temperature$

$T_{0} = Initial \ temperature \ of \ the \ body$

$k = constant$

From the question,

$T_{0} = 86 ^{o}C$

$T_{s} = -20 ^{o}C$

∴ $T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C$

$T_{0} - T_{s} = 106^{o} C$

Therefore, the equation $T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$ becomes

$T_{(t)}=-20+106 e^{kt}$

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time $t = 1 \ hour$ , $T_{(t)} = 58^{o}C$

Then, we can write that

$T_{(1)}=58 = -20+106 e^{k(1)}$

Then, we get

$58 = -20+106 e^{k(1)}$

Now, solve for $k$

First collect like terms

$58 +20 = 106 e^{k}$

$78 =106 e^{k}$

Then,

$e^{k} = \frac{78}{106}$

$e^{k} = 0.735849$

Now, take the natural log of both sides

$ln(e^{k}) =ln( 0.735849)$

$k = -0.30673$

This is the value of the constant $k$

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at $t = 2 \ hours$

From

$T_{(t)}=-20+106 e^{kt}$

Then

$T_{(2)}=-20+106 e^{(-0.30673 \times 2)}$

$T_{(2)}=-20+106 e^{-0.61346}$

$T_{(2)}=-20+106\times 0.5414741237$

$T_{(2)}=-20+57.396257$

$T_{(2)}=37.396257 \ ^{o}C$

$T_{(2)} \approxeq 37.40 \ ^{o}C$

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

Learn more here: brainly.com/question/11689670

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2 years ago

Read 2 more answers

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ValentinkaMS [17]

Answer:

Step-by-step explanation:

Good luck

4 0

3 years ago