Using the function below: {(-5,2),(1,1),(5,1),(2,6)} f(1) = f(2)=

1 answer:

4 0

$\bf \{(-5,2),(1,1),(5,1),(2,6)\}\implies \begin{array}{|cc|ll} \cline{1-2} x&f(x)\\ \cline{1-2} -5&2\\ 1&1\\ 5&1\\ 2&6\\ \cline{1-2} \end{array} \begin{array}{llll} \\[1em] \leftarrow f(1) = 1\\\\ \leftarrow f(2)=6 \end{array}$

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[02.05] Marcie solved the following inequality, and her work is shown below: −2(x − 5) − 12 ≤ 4 + 6(x + 3) −2x + 10 − 12 ≤ 4 + 6

Nezavi [6.7K]

Hi there! Marcie made a mistake in the last step of the solving process.

Let's solve this problem step by step!
−2(x − 5) − 12 ≤ 4 + 6(x + 3)

First work out the parenthesis, which can for instance be done using rainbow method.

−2x + 10 − 12 ≤ 4 + 6x + 18
The parenthesis are worked out correctly. Remember that multiplying a negative by a negative gives you a positive.

Now we can collect the terms.
−2x − 2 ≤ 6x + 22

The terms are collected correctly.

Next up is subtracting 6x from both sides and adding 2 to both sides of the inequality.
−8x ≤ 24

This step is also correct.
Finally we must divide both sides of the inequality by -8 to find our answer. Be aware of the fact that the sign flips when dividing by a negative!

x ≤ −3
Mistake! Marcie forgot to flip the sign. Instead the answer should be $x \geqslant - 3$

~ Hope this helps you!

4 0

4 years ago

A student missed 3 questions on a 30 question test. what percent did she get correct?

Lyrx [107]

Answer:

90%

Explanation: 27/30 ÷ 3 = 9/10

9/10 ⇒ 9 x 10 = 90%

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3 years ago

The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter and area in square units

ad-work [718]

Check the picture below.

the triangle has that base and that height, recall that A = 1/2 bh.

now as for the perimeter, you can pretty much count the units off the grid for the segment CB, so let's just find the lengths of AC and AB,

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&C&(~ 6 &,& 2~)
\end{array}~~~ 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{(6-2)^2+(2-8)^2}\implies AC=\sqrt{4^2+(-6)^2}
\\\\\\
AC=\sqrt{16+36}\implies AC=\sqrt{52}\implies AC=\sqrt{4\cdot 13}
\\\\\\
AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&B&(~ 16 &,& 2~)
\end{array}\\\\\\
AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2}
\\\\\\
AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58}
\\\\\\
AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}$

so, add AC + AB + CB, and that's the perimeter of the triangle.