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MaRussiya [10]
3 years ago
7

A school ordered 32 orange shirts, 29 blue shirts, and 34 green shirts. The shirts are being divided equally among 6 classrooms.

If each classroom gets the same number of shirts, how many shirts will be left over?
A.
2
B.
4
C.
5
D.
3
Mathematics
2 answers:
Ratling [72]3 years ago
8 0

Answer: D

Step-to-step explanation:

belka [17]3 years ago
6 0

Answer:

C

Step-by-step explanation:

The orange shirts will have 32/6 = 5 with 2 left over

The blue shirts will have 29/6 = 4 with 5 left over

The green shirts will have 34/6 = 5 with 4 left over

The left overs are 2 + 5 + 4 = 11 but I have divided the colors equally. If color does not matter then the total number of shirts = 32 + 29  + 34

The total number of shirts is 95

95/6 = 15 with a remainder of 5

So 5 shirts will be left over.

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Hope that helps!

3 0
3 years ago
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An ice cream shop sells cones at the volume of 94.2 cubic meters they want to double the volume of the cones without changing th
marin [14]
The volume of a cone is V=\pi r^2\frac{h}3 where r = radius and h = height. If the cone has a volume of 94.2 cm³ (I assume you didn't mean m³ because that would be ridiculously huge) and a height of 10 cm, we can plug these values into the formula to find the radius. Don't do any rounding.

94.2 = \pi r^2\frac{10}3 \\ 282.6 = \pi r^2 *10 \\ 28.26 = \pi r^2 \\ 8.99543738355 = r^2 \\ 2.99923946752=r

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5 0
3 years ago
The proof shows that ABCD is a rectangle. Which of the following is the missing reason?
patriot [66]

Answer:

Side-side-side postulate

Step-by-step explanation:

In the first statement, we are given that AC = BD ....... (1){It is given}

Now, in the second statement, we are given that AB = DC ...... (2){ Because the parallelogram has equal opposite sides}

Now, again we are given that AD = DA ........... (3) {From the symmetric property}

Now, conditions (1), (2), and (3) are applicable to say that Δ ABD and Δ ACD are congruent.

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AveGali [126]

\huge\mathscr{\colorbox{blue}{\color{yellow}{Answer:}}}

\large{\rightarrow{6\:(x\:+\:4)\:=\:-\:2\:(4x\:-\:4)\:+\:8x}}

\large{\rightarrow{6\:(x\:+\:4)\:=\:-\:2\:(4x\:-\:4)\:+\:8x}}

\large{\rightarrow{6x\:+\:24\:=\:-\:8x\:+\:8\:+\:8x}}

\large{\rightarrow{6x\:+\:24\:=\:{\cancel{-\:8x}}\:+\:8\:+{\cancel{\:8x}}}}

\large{\rightarrow{6x\:+\:24\:=\:8}}

\large{\rightarrow{6x\:=\:8\:-\:24}}

\large{\rightarrow{6x\:=\:16}}

\large{\rightarrow{x\:=\:\frac{-\:16}{6}}}\\

\large\pink{\boxed{\color{lime}{\rightarrow{x\:=\:\frac{-\:8}{3}}}}}\\

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