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3 years ago

How is jimthompson so smart?

Mathematics

1 answer:

Charra [1.4K]3 years ago

6 0

The statement is valid. To set up a Euler diagram, follow these steps

Step 1) Draw a large circle and label it circle A

Step 2) Inside this large circle, draw a smaller circle and label it circle B

Circle A will represent the set of all obtuse angles (any angles that are between 90 and 180 degrees). Circle B represents only one item: the angle value of 150 degrees. All of circle B is inside circle A because 150 degrees is between 90 and 180; therefore, 150 is obtuse. This is simply the definition of what "obtuse" means.

So if you threw a dart and it landed in circle B, then you know for sure it landed in circle A as well. This is effectively saying "if you get a 150 degree angle, then you know its obtuse"

This is a visual way to back up the conditional statements you used earlier and help add more evidence that the statement "A 150 degree angle is an obtuse angle" is a valid statement.

You might be interested in

Using the digits 0 to 9

Gre4nikov [31]

Supplementary Angles:

$\boxed{100°} and \boxed{80°}$

Arithmetically speaking, the closest 2 supplementary angles can get (in 3 and 2 digits respectively) is the upwritten.

Complementary Angles:

$\boxed{45°} and \boxed{45°}$

Simply, in this case, for angles to be numerically as close as possible - make both the angles 45°.

8 0

2 years ago

What’s the radius of this button ?

Neporo4naja [7]

Answer:

2.5 cm

Step-by-step explanation:

We see the diameter is 5cm

so the radius is half the diameter

5/2 = 2.5 cm

3 0

4 years ago

Read 2 more answers

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$