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elena-s [515]
3 years ago
5

Simplify the expression. 0.9 - 2(3r + 0.2)

Mathematics
2 answers:
kenny6666 [7]3 years ago
8 0
The simplified expression is 0.5-6r
I am Lyosha [343]3 years ago
3 0

Answer:

-6r + .5

reeeeeeeeeeeeeeeeerrrrreeeeeeeeeeeere

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Makovka662 [10]

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Step-by-step explanation:

6 0
3 years ago
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The bake star bakery uses 2 1/4 cups of raisins to make 4 servings of trail mix. How many cups of raisins are in each serving?
Dima020 [189]

let's firstly convert the mixed fraction to improper fraction, and then divide it by 4 to see what our quotient is.


\bf \stackrel{mixed}{2\frac{1}{4}}\implies \cfrac{2\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{9}{4}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{9}{4}\div 4\implies \cfrac{9}{4}\div \cfrac{4}{1}\implies \cfrac{9}{4}\cdot \cfrac{1}{4}\implies \cfrac{9}{16}

3 0
3 years ago
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What is the area of trapezoid ABCD ? (Enter your answer as a decimal or whole number)
MrRa [10]
Check the picture below.

bear in mind that, the "bases" are the two parallel sides, and the height is the distance between them.

\bf \textit{area of this trapezoid}\\\\
A=\cfrac{AB(BC+AD)}{2}\\\\
-------------------------------\\\\
\textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
A&({{ -1}}\quad ,&{{ 5}})\quad 
%  (c,d
B&({{ 3}}\quad ,&{{ 2}})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
AB=\sqrt{[3-(-1)]^2+[2-5]^2}\implies AB=\sqrt{(3+1)^2+(2-5)^2}
\\\\\\
AB=\sqrt{16+9}\implies AB=\sqrt{25}\implies \boxed{AB=5}

\bf -------------------------------\\\\
\textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
B&({{ 3}}\quad ,&{{ 2}})\quad 
%  (c,d
C&({{ 0}}\quad ,&{{ -2}})
\end{array}
\\\\\\
BC=\sqrt{(0-3)^2+(-2-2)^2}\implies BC=\sqrt{9+16}
\\\\\\
BC=\sqrt{25}\implies \boxed{BC=5}

\bf -------------------------------\\\\
\textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
A&({{ -1}}\quad ,&{{ 5}})\quad 
%  (c,d
D&({{ -13}}\quad ,&{{ -11}})
\end{array}
\\\\\\
AD=\sqrt{[-13-(-1)]^2+[-11-5]^2}
\\\\\\
AD=\sqrt{(-13+1)^2+(-16)^2}\implies AD=\sqrt{144+256}
\\\\\\
AD=\sqrt{400}\implies \boxed{AD=\sqrt{20}}

so, the area for this trapezoid is then

\bf A=\cfrac{5(5+20)}{2}\implies A=\cfrac{125}{2}\implies A=62\frac{1}{2}

8 0
4 years ago
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blagie [28]
Okay, so to find like denominators you need to find a number that both denominators go into. For 15, they both go into 45. 9 goes in 5 times and 15 goes in 3 times. So multiply their numerators by 5 and 3. It would be 25/45 for 5/9 and it would be 12/45 for 4/15. Do the same for 16 and 17, for 16 they both go into 42 and for 17 they both go into 42.
7 0
4 years ago
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