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antiseptic1488 [7]
3 years ago
9

Simplify the expression below. (6-3)(66)

Mathematics
1 answer:
irinina [24]3 years ago
6 0

Answer:

198

Step-by-step explanation:

(6-3) x 66 subtract

3x66 multiply

198

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Of the 30 students in the sixth period math class, 8 are also in the same fourth period science class. Which can be used to dete
Kazeer [188]

Answer:

A. (8/30)(22/29)(21/28)

Step-by-step explanation:

The key is in reading the question really thoroughly, like, way more than you would expect. The <em>first</em> person is from the group of 8, thus (8/30), but the other <em>two</em> <u>cannot</u> be from the group of 8. 30(students total) minus 8 (the 8 people from science class)=22 (remaining in math). You choose one person who is <em>not</em> from the group of 8, and you are left with 21 other people. The number of students available to be in a group goes down each time: 30 down to 29 down to 28 (the denominators). Kaboom, A. (8/30)(22/29)(21/28).

8 0
3 years ago
Read 2 more answers
How do you solve: 4 + |k + 1| &lt;15
Viktor [21]
4+|k+1|<15
-4           -4      15-4=11
|k+1|<11         |k+1|=k+1
k+1<11
  -1    -1
k<10
6 0
3 years ago
Ava has 34 candy bars. If
aliya0001 [1]

\bf Step-by-step~explanation:

If Ava has 34 candy bars, and each box can hold 5 bars, then we need to find out how many boxes that are filled up.

\bf Step~1:

Divide the number of candy bars (34), by the number each box can hold (5)

\bf\frac{34}{5}=6.8

Since we cannot have 6.8 boxes, we have to round down to 6.

\boxed{\bf Our~final~answer:~6~boxes}

\bf Check~our~answer:

To check our answer, we multiply the number of boxes (6), by the number of bars in each box (5), to get 30. We add Ava's extra bars (4), and we get the number we started off with: 34. This proves our answer is correct!

4 0
3 years ago
Uh yeah so this again lol
Andrej [43]

Answer:

follow the insructions from the answer before

Step-by-step explanation:

3 0
3 years ago
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A company manufactures aluminum mailboxes in the shape of a box with a half-cylinder top. The company will make 1485 mailboxes t
Gnoma [55]

Answer:8746.65\ m^2

Step-by-step explanation:

Given

Base of mail box is 0.4\ m\times 0.7\ m

height of mail box h=0.55\ m

Total area required for one mail box

A_t=\text{surface area of cubiod} - \text{Top of cubiod +Semi-cylinder area}

A_t=2(lb+bh+hl)-lb+\pi rh'

where b=0.4\ m

l=0.7\ m

h=0.55\ m

r=\frac{0.4}{2}=0.2\ m

h'=0.7\ m

Substituting values

A_t=2(0.7\times 0.4+0.4\times 0.55+0.55\times 0.7)-0.4\times 0.7+\pi \times 0.2\times 0.7

A_t=2(0.28+0.385+0.22)-0.28+\pi \times 0.14

A_t=1.77-0.28+4.4

A_t=5.89\ m^2

Therefore for 1485 mailboxes Area required

A_o=1485\times 5.89=8746.65\ m^2

4 0
4 years ago
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