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Cloud [144]
3 years ago
12

What is the coefficient of 2x^2+5x^3+4x+3?​

Mathematics
1 answer:
ki77a [65]3 years ago
6 0
The leading coefficient of 2x^2+5x^3+4x+3 is 5
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PLEAAASE HELP! i will manifest good things for you. i will mark u brainliest LOL
melamori03 [73]

Answer:

(c) is m=1 i am working on the other ones

5 0
3 years ago
A gardener measures every plant in her garden and notes that 28% are under 6 inches, 54% are between six and 8 inches and the re
Masteriza [31]

<u>Answer:</u>

The vertical axis should begin with 0 inches.

<u>Step-by-step explanation:</u>

The vertical axis must start with 0 inches to ensure accuracy for the measures of every plant since 28% of the plants in the garden are under 6 inches.

If the vertical axis begins with 6 inches or above, we can miss a lot of the gardener's data, especially the plants which are under 6 inches.

Therefore, to ensure that the graph does not mislead any measures of the plant, the vertical axis must begin with 0 inches.

3 0
3 years ago
Read 2 more answers
A. 112 cm2<br><br> B. 64 cm2<br><br> C. 48 cm2<br><br> D. 56 cm2
goldfiish [28.3K]
The area of a triangle is taken by multiplying the length of the base times the height then dividing by 2.
l = 14cm
h = 8cm
A = \frac{l * h}{2}
A = \frac{14 * 8}{2}
A = \frac{112}{2}
A = 56 cm^{2}
6 0
4 years ago
A sport arena covers 710,430 square feet of ground a news paper reported that the arena covers about 700,000 square feet of grou
n200080 [17]
700,000 is your final answer so the number was rounded to the hundred thousands value/place.
4 0
3 years ago
can you transform the square shown below into pieces that when put together form five small equal squares whose total area shall
Sunny_sXe [5.5K]

How about this (see attached image):

Use the four cuts as shown in the image (red lines).

Then assemble 5 equal squares by the numbers: 1 center square and the rest are pieced together using two pieces as shown. All five together add up to the same area as original square because we use all pieces.

The way one gets a hint toward a solution is to see how an area of a square of length 1 can be split into 5 equal square areas:

1^2 = 5\cdot x^2\\x = \frac{\sqrt{5}}{5} = \sqrt{\frac{2^2}{5^2}+\frac{1^2}{5^2}}

which indicates we need to find a a triangle with sides 2 and 1 to get the hypotenuse of the right length. That gave rise to the cut pattern (if you look carefully, there are triangles with those side lengths).

6 0
3 years ago
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