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ehidna [41]
3 years ago
8

What is all the answers to a 2 × problem​

Mathematics
2 answers:
nalin [4]3 years ago
8 0
Double the number like 2 x5 would be 5+5
monitta3 years ago
3 0

Answer:

you double the number

Step-by-step explanation:

example 2 x 3 = 6

You might be interested in
Find the perimeter. Simplify your answer.
dimaraw [331]

Answer:

30y+14

Step-by-step explanation:

add the values of all the sides together: (5y+4) + (5y+4) + (10y+3) + (10y+3) = (5y + 5y + 10y + 10y) + (4 + 4 + 3 + 3)= 30y + 14

5 0
2 years ago
If we inscribe a circle such that it is touching all six corners of a regular hexagon of side 10 inches, what is the area of the
Brrunno [24]

Answer:

\left(100\pi - 150\sqrt{3}\right) square inches.

Step-by-step explanation:

<h3>Area of the Inscribed Hexagon</h3>

Refer to the first diagram attached. This inscribed regular hexagon can be split into six equilateral triangles. The length of each side of these triangle will be 10 inches (same as the length of each side of the regular hexagon.)

Refer to the second attachment for one of these equilateral triangles.

Let segment \sf CH be a height on side \sf AB. Since this triangle is equilateral, the size of each internal angle will be \sf 60^\circ. The length of segment

\displaystyle 10\, \sin\left(60^\circ\right) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}.

The area (in square inches) of this equilateral triangle will be:

\begin{aligned}&\frac{1}{2} \times \text{Base} \times\text{Height} \\ &= \frac{1}{2} \times 10 \times 5\sqrt{3}= 25\sqrt{3} \end{aligned}.

Note that the inscribed hexagon in this question is made up of six equilateral triangles like this one. Therefore, the area (in square inches) of this hexagon will be:

\displaystyle 6 \times 25\sqrt{3} = 150\sqrt{3}.

<h3>Area of of the circle that is not covered</h3>

Refer to the first diagram. The length of each side of these equilateral triangles is the same as the radius of the circle. Since the length of one such side is 10 inches, the radius of this circle will also be 10 inches.

The area (in square inches) of a circle of radius 10 inches is:

\pi \times (\text{radius})^2 = \pi \times 10^2 = 100\pi.

The area (in square inches) of the circle that the hexagon did not cover would be:

\begin{aligned}&\text{Area of circle} - \text{Area of hexagon} \\ &= 100\pi - 150\sqrt{3}\end{aligned}.

3 0
3 years ago
X2 + x - 132 = 0<br> what is the formula
geniusboy [140]
3x-132=0
you need to add all like terms first then answer the equation
4 0
3 years ago
Read 2 more answers
Check
8_murik_8 [283]

Answer:

The answer to your question is given below.

Step-by-step explanation:

To which of the above expression is a sum or difference of cube, or not a sum or difference of cube, we shall do the following simplification:

Note: The Cube root of a particular number is simply a multiplication of an identical number in three places.

64x³ – 216

64 has a cube root of 4 and 216 has a cube root of 6. Therefore, the above expression can be written as:

4³x³ – 6³

(4x)³ – 6³

64x³ – 216 = (4x)³ – 6³

Therefore, 64x³ – 216 can be expressed as a difference of cube.

8x^9 + 27

8 has a cube root of 2, x^9 has a cube root of x³ and 27 has a

cube root of 3. Therefore, the above expression can be written as:

2³(x³)³ + 3³

(2x³)³ + 3³

8x^9 + 27 = (2x³)³ + 3³

8x^9 + 27 can be expreessed as a sum of cube

x³ + 125

125 has a cube root of 5. Therefore, the above expression can be written as:

x³ + 5³

x³ + 125 = x³ + 5³

x³ + 125 can be expressed as a sum of cube

36x³ + 121

36 and 121 has no cube root. Therefore, the above expression is not a sum or difference of cube.

x^6 – 16

x^6 has a cube root of x² and 16 has no cube root. Therefore, the above expression is not a sum or difference of cube.

Summary:

Sum or Difference of cubes

64x³ – 216

8x^9 + 27

x³ + 125

Not a Sum or Difference of cubes

36x³ + 121

x^6 – 16

5 0
3 years ago
At a school cafeteria a cold lunch costs $1.80, and a hot lunch costs $3.00. During one school year a teacher spent
svetoff [14.1K]
160 is cold rest is hot c:&:&:
7 0
3 years ago
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