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I am Lyosha [343]
4 years ago
8

Ken’s predicted temperature for April is 7ºC. His prediction for May is 2ºC higher than April. He used the number line to solve

the problem, starting at 0, moving right to 7, and then moving 2 units to the left. Explain Ken’s error
Mathematics
2 answers:
Gekata [30.6K]4 years ago
7 0

Answer:

He should have moved to the right.

Step-by-step explanation:

Ken moved 7 to the right which is correct because he needs to get 7º higher, but when he moved 2º to the left that would be a temperature decrease which is is incorrect.

Masteriza [31]4 years ago
4 0

Answer:

Sample Response: Ken’s error was showing a decrease in temperature by moving 2 units left on the number line. He should have moved 2 units right to represent an increase in temperature of 2°C.

Step-by-step explanation:

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Step-by-step explanation:

3 0
3 years ago
Harry ,Louis and Niall Are working with exponents. Harry claims 4^2•4^5=4^10 Louis claims 4^2•4^5=4^7 Niall claims 4^2•4^5=16^7
slavikrds [6]

Answer:

Louis is correct

Step-by-step explanation:

we know that

When multiplying two powers that have the same base, you can add the exponents

so

a^{n} a^{m}=a^{n+m}

so

In this problem we have

4^{2} 4^{5}=4^{2+5}=4^{7}

therefore

Louis is correct

4 0
4 years ago
Two cars traveled equal distances in different amounts of time. Car A traveled the distance in 2.4 h, and Car B traveled the dis
Andrej [43]
Let the distance be D miles
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7 0
3 years ago
Read 2 more answers
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
Please help me solve!!
larisa [96]
The answer is A just use the function

5 0
3 years ago
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