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Katena32 [7]
3 years ago
9

What is the constant term in the expansion of the binomial (x − 2)^4?

Mathematics
1 answer:
Marianna [84]3 years ago
3 0

Answer:

the answer is a i think





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someone please help me with this if one person already answers then dont answer i only have a few points left and need it for sc
Viefleur [7K]
The answer should be -3


Hope this helps :)
7 0
3 years ago
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as a gift, you fill the calendar with packets of chocolate candy. Each packet has a volume of 2 cubic inches. Find the maximum n
qaws [65]
The picture of the calendar is shown in the attached image.

Now, first we will get the volume of the calendar itself, we can note the calendar has the shape of a triangular prism.
Volume of triangular prism = area of base * depth
The area of base = area of triangle = 1/2 * base * depth 
Therefore:
Volume of prism = 1/2 * base * height * depth
where:
base = 4 in
height is he height of the base = 6 in
depth is the depth of the calendar = 8 in
Therefore:
Volume of calendar = 1/2 * 4 * 6 * 8 = 96 in^3

Now, we are given that the volume of each candy is 2 in^3, this means that:
number of candies to fill the calendar = volume of calendar / volume of candy
                                                            = 96/2 
                                                            = 48 candies

Hope this helps :)

4 0
4 years ago
In front of a store, there is a row of parking spaces. Cars park parallel to one another, with the front of each car facing the
Gelneren [198K]

The width used for the car spaces are taken as a multiples of the width of

the compact car spaces.

Correct response:

  • The store owners are incorrect
<h3 /><h3>Methods used to obtain the above response</h3>

Let <em>x</em><em> </em>represent the width of the cars parked compact, and let a·x represent the width of cars parked in full size spaces.

We have;

Initial space occupied = 10·x + 12·(a·x) = x·(10 + 12·a)

New space design = 16·x + 9×(a·x) = x·(16 + 9·a)

When the dimensions of the initial and new arrangement are equal, we have;

10 + 12·a = 16 + 9·a

12·a - 9·a = 16 - 10 = 6

3·a = 6

a = 6 ÷ 3 = 2

a = 2

Whereby the factor <em>a</em> < 2, such that the width of the full size space is less than twice the width of the compact spaces, by testing, we have;

10 + 12·a < 16 + 9·a

Which gives;

x·(10 + 12·a) < x·(16 + 9·a)

Therefore;

The initial total car park space is less than the space required for 16

compact spaces and 9 full size spaces, therefore; the store owners are

incorrect.

Learn more about writing expressions here:

brainly.com/question/551090

5 0
2 years ago
CAN SOMEONE HELP ME PLEASE???
lana66690 [7]

Answer:

Step-by-step explanation: First, you need to write all your data down from least to greatest. Next, count up all the numbers to see how many students. Then you want to find the mean which is average (add then divide). Median is the middle so you will x each number out till you get to the middle, and if the middle is 2 numbers, you find the average of them to find the median. Mode means most often. Range is highest minus lowest. And you can do the next 2.

4 0
3 years ago
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There are 35 counters in a bag.
horsena [70]

Answer:

debes hacer l regla de tres

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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