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PilotLPTM [1.2K]
3 years ago
14

HELP I NEED HELP! Can someone tell me how is it that answer I DO NOT WANT ONLY THE LETTER. I also want to know why is it that an

swer. Plz Help.
Mathematics
1 answer:
Strike441 [17]3 years ago
6 0
Sure what is the question? like what needs to be figured out?
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What is the product? 7⋅(−4)⋅(−6) −168 −3 17 168
Lunna [17]

Answer:

The Product of 7⋅(−4)⋅(−6) −168 −3 is<em> </em><u>-3</u>

8 0
3 years ago
Hey please help will mark brainliest / please give a genuine anwser
Bad White [126]

Answer:

Shaded area is 392

Step-by-step explanation:

The triangles on the left have heights 14 and 14, the ones on the right have 20 and 8.

The bases are equal, so we can compute them as (40-12)/2 = 28/2 = 14

The triangles have areas:

14*14/2 = 14*7 = 98

14*14/2 = 14*7 = 98

20*14/2 = 20*7 = 140

8*14/2 = 8*7 = 56

note that the triangles on the left and the ones on the right have the same total area! That would have happened no matter what the height split was!

The total area is 98+98+140+56 = 196 + 196 = 392

7 0
3 years ago
What is the circumference of a dinner plate with a radious of 4.5 inches​
____ [38]

Answer:

28.27 inches

Step-by-step explanation:

-Circumference of a circle is calculated using the formula:

C=\pi D\\\\=2\pi r

#Substitute for radius in the formula to solve for C:

C=2\pi r\\\\=2\pi \times 4.5 \\\\=28.27 \ in

Hence, the plate's circumference is 28.27 inches

4 0
3 years ago
The larger of two numbers is five more than twice the smaller. The difference is thirteen. Find numbers.
KIM [24]
The numbers are x and y
x>y

x=5+2y

x-y=13
so
sub 5+2y for x
5+2y-y=13
5+y=13
y=8

sub back
x-y=13
x-8=13
21=x

the numbres are 21 and 8
4 0
3 years ago
How many ways are there to arrange the first five letters of the alphabet?
Nonamiya [84]
In probability, problems involving arrangements are called combinations or permutations. The difference between both is the order or repetition. If you want to arrange the letters regardless of the order and that there must be no repetition, that is combination. Otherwise, it is permutation. Therefore, the problem of arrange A, B, C, D, and E is a combination problem.

In combination, the number of ways of arranging 'r' items out of 'n' items is determined using n!/r!(n-r)!. In this case, you want to arrange all 5 letters. So, r=n=5. Therefore, 5!/5!(505)! = 5!/0!=5!/1. It is simply equal to 5! or 120 ways.
6 0
3 years ago
Read 2 more answers
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