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mihalych1998 [28]
3 years ago
10

Please help: The solution to which inequality is shown? A. y + 5 > –2 B. y + 5 ≥ –2 C. y + 5 < –2 D. y + 5 ≤ –2

Mathematics
1 answer:
yKpoI14uk [10]3 years ago
8 0
You see that the first circle is between -10 and -5. This means that part of the inequality must be between these two values. In addition, the open circle means that the inequality is either less than or greater than, but not equal to. This rules out both B and D. Since the arrow is pointing to the right, this means that the inequality is greater than. Now let's look for the inequality with a value greater than a value between -10 and -5. 

A.
y+5>-2
y>-7

C.
y+5<-2
y<-7

The one with greater than is A so that is your answer.

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In the formula, C = (F - 32) * 5/9, the letter "F" is known as which of the following?
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A recipe for a cake instructs you to bake it at 220° C for 45 minutes. How many degrees Fahrenheit is this?
sergiy2304 [10]

This is an exercise on Thermometric scales.

We have as data:

  • T = 220 °C
  • T =  °F ??

We apply the following formula:

\large\displaystyle\text{$\begin{gathered}\sf ^{\circ}F=\frac{9}{5}\  ^{\circ}C+32  \end{gathered}$}

We substitute data in the formula:

\large\displaystyle\text{$\begin{gathered}\sf ^{\circ}F=\frac{9}{5}\times220+32  \end{gathered}$}

First multiply 9 x 220, then the result of this division is divided by 5.

\large\displaystyle\text{$\begin{gathered}\sf =396+32 \ \ \to \ \ [Add] \end{gathered}$}

\boxed{\large\displaystyle\text{$\begin{gathered}\sf =428 \ ^{\circ}F \end{gathered}$}}

Therefore the cake recipe that should be baked at 220 °C for 45 minutes, on the Fahrenteir scale is 428 °F. Which indicates that it will be very hot.

3 0
2 years ago
What are all the potential rational roots of f(x)=15x^11 -6x^8 +x^3-4x+3
Lady_Fox [76]

The theorem of rational solutions states that, given a polynomial with integer coefficients, every potenial rational solution is written as \frac{p}{q}, where p divides the constant term, and q divides the leading term.


So, our choices for p are \pm 1, \pm 3, while our choices for q are \pm 1, \pm 3, \pm 5, \pm 15.


Combine all possible numerator and denominator to get all the potenial rational solutions:


\pm\frac{1}{1},\pm\frac{1}{3},\pm\frac{1}{5},\pm\frac{1}{15}


\pm\frac{3}{1},\pm\frac{3}{3},\pm\frac{3}{5},\pm\frac{3}{15}

4 0
4 years ago
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