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Helga [31]
3 years ago
15

Is it is it possible to have 43 pencils and divide the pencils into groups with the same number of pencils with none left over

Mathematics
2 answers:
anastassius [24]3 years ago
6 0

we have given 43 pencils

we need to divide it equally so that none can be without groups.

we need to find the factors of 43.

we know that 43 is a prime number. which cannot be divided.so it is not possible to divide the 43 pencils with none left over .

sladkih [1.3K]3 years ago
5 0

TO find out the answer, we need to check whether 43 have any factors or not

43 is a prime number that is it have only two factors 1 and 43 . So either 1 will take whole 43 pencils or 43 will take one pencil each .

So it is not possible to divide 43 pencils into groups with the same number of pencils and none left over .

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VARVARA [1.3K]

Answer:

1.8

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

Step-by-step explanation:

<u>Step 1: Define</u>

2.8 - 4 ÷ 4

<u>Step 2: Evaluate</u>

  1. Division:                                                                                                             2.8 - 1
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8 0
3 years ago
write the standard form of the equation of the circle for which the endpoints of a diameter are (0,0) and (4,-6)
Elenna [48]

Given:

The endpoints of a diameter are (0,0) and (4,-6).

To find:

The equation of the circle.

Solution:

The endpoints of a diameter are (0,0) and (4,-6). So, the length of the diameter is

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

d=\sqrt{(-6-0)^2+(4-0)^2}

d=\sqrt{36+16}

d=\sqrt{52}

d=2\sqrt{13}

Now, radius is half of the diameter.

r=\dfrac{2\sqrt{13}}{2}

r=\sqrt{13}

Center of the circle is the midpoint of the endpoints of a diameter.

Center=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)

Center=\left(\dfrac{0+4}{2},\dfrac{0+(-6)}{2}\right)

Center=\left(\dfrac{4}{2},\dfrac{-6}{2}\right)

Center=\left(2,-3\right)

Standard form of a circle is

(x-h)^2+(y-k)^2=r^2

where, (h,k) is center and r is radius.

The center of the circle is (2,-3) and radius is \sqrt{13}. So,

(x-2)^2+(y-(-3))^2=(\sqrt{13})^2

(x-2)^2+(y+3)^2=13

Therefore, the standard form of the circle is (x-2)^2+(y+3)^2=13.

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<u></u>

<u>Length is 12 and width is 10</u>

<u></u>

<u>check if its correct </u>

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