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Zielflug [23.3K]
4 years ago
11

For a scientific Experiment, A physicist must make sure that the temperature of a metal at 0°C get to no colder then -80°C. the

physicist changes the metals temperature at a steady rate of -4°C per hour.for how long can the physicist change the temperature
Mathematics
1 answer:
shepuryov [24]4 years ago
8 0
80 divided by 4=20. ANSWER is 20 hours
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HELP please!!!!! I'm honestly dying I hate math
Sveta_85 [38]
2) 100
3) 49
4) 19
5) idk
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7) 7
3 0
3 years ago
A bookstore charges a standard rate for paperback and hardback bestseller books. The cost of each paperback book is $8 less than
11111nata11111 [884]

Answer:

Hardback- $328 , Paperback- $1548

5 0
3 years ago
Read 2 more answers
Suppose a and b are the solutions to the quadratic equation 2x^2-3x-6=0. Find the value of (a+2)(b+2).
KonstantinChe [14]

Answer:

(a+2)(b+2) = 4

Step-by-step explanation:

We are given the following quadratic equation:

2x^2-3x-6=0

Let a a and b be the solution of the given quadratic equation.

Solving the equation:

2x^2-3x-6=0\\\text{Using the quadratic formula}\\\\x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\text{Comparing the equation to }ax^2 + bx + c = 0\\\text{We have}\\a = 2\\b = -3\\c = -6\\x = \dfrac{3\pm \sqrt{9-4(2)(-6)}}{4} = \dfrac{3\pm \sqrt{57}}{4}\\\\a = \dfrac{3+\sqrt{57}}{4}, b = \dfrac{3-\sqrt{57}}{4}

We have to find the value of (a+2)(b+2).

Putting the values:

(a+2)(b+2)\\\\=\bigg(\dfrac{3+\sqrt{57}}{4}+2\bigg)\bigg(\dfrac{3-\sqrt{57}}{4}+2\bigg)\\\\=\bigg(\dfrac{11+\sqrt{57}}{4}\bigg)\bigg(\dfrac{11-\sqrt{57}}{4}\bigg)\\\\=\dfrac{121-57}{156} = \dfrac{64}{16} = 4

3 0
3 years ago
Find the product of (p+5)(p-2)
Korolek [52]

Answer:

The answer is p^2+3p-10

5 0
3 years ago
What is the range of the following function:<br> y = 2(5^x) - 1
Orlov [11]
<h2>Hello!</h2>

The answer is:

The range of the function is:

Range: y>2

or

Range: (2,∞+)

<h2>Why?</h2>

To calculate the range of the following function (exponential function) we need to perform the following steps:

First: Find the value of "x"

So, finding "x" we have:

y=2(5^{x}-1)\\\frac{y}{2}=5^{x}-1\\\\\frac{y}{2}-1=5^{x}\\\\Log_{5}(\frac{y}{2}-1)=Log_5(5^{x})\\\\x=Log_{5}(\frac{y}{2}-1)

Second: Interpret the restriction of the function:

Since we are working with logarithms, we know that the only restriction that we found is that the logarithmic functions exist only from 0 to the possitive infinite without considering the number 1.

So, we can see that if the variable "x" is a real number, "y" must be greater than 2 because if it's equal to 2 the expression inside the logarithm will tend to 0, and since the logarithm of 0 does not exist in the real numbers, the variable "x" would not be equal to a real number.

Hence, the range of the function is:

Range: y>2

or

Range: (2,∞+)

Note: I have attached a picture (the graph of the function) for better understanding.

Have a nice day!

5 0
3 years ago
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