Just cross multiply basically divide
Answer: 7 i think
Step-by-step explanation:
I am a number greater than 40,000 and less than 60,000:
40,000 < n < 60,000
This means that:
n = 10,000n₁ + 1,000n₂ + 100n₃ + 11n₄
And also:
4 ≤ n₁ < 6
0 ≤ n₂ ≤ 9
0 ≤ n₃ ≤ 9
0 ≤ n₄ ≤ 9
My ten thousands digit is 1 less than 3 times the sum of my ones digit and tens digit:
n₁ = 3*2n₄ - 1
n₁ = 6n₄ - 1
This means that:
n = 10,000*(6n₄-1) + 1,000n₂ + 100n₃ + 11n₄
n = 60,000n₄ - 10,000 + 1,000n₂ + 100n₃ + 11n₄
n = 60,011n₄ - 10,000 + 1,000n₂ + 100n₃
<span>My thousands digit is half my hundreds digit, and the sum of those two digits is 9:
n</span>₂ = 1/2 * n₃
<span>
n</span>₂ + n₃ = 9
<span>
Therefore:
n</span>₂ = 9 - n₃
<span>
Therefore:
9 - n</span>₃ = 1/2 * n₃
<span>
9 = 1/2 * n</span>₃ + n₃
<span>
9 = 1.5 * n</span>₃
<span>
Therefore:
n</span>₃ = 6
<span>
If n</span>₃=6, n₂=3.
<span>
This means that:
</span>n = 60,011n₄ - 10,000 + 1,000*3 + 100*6
n = 60,011n₄ - 10,000 + 3,000 + 600
n = 60,011n₄ - 6,400
Therefore:
0<n₄<2, so n₄=1.
If n₄=1:
n = 60,011 - 6,400
n = 53,611
Answer:
53,611
Answer:
42.43
Step-by-step explanation:
Given:
A circle with a radius of 5 cm sits inside a 11 cm x 11 cm rectangle.
Question asked:
What is the area of the shaded region as shown in the figure ?
Solution:
First of all calculate area of circle and then area of rectangle:-


<u>As here length an width both are 11 cm, we can say that this is a square.</u>
<u />

Now, area of shaded region = Area of square - Area of circle
= 121 - 78.571 = 42.429 
Therefore, the area of the shaded region will be 42.43 
Ah...Trigonometry is fun!
The law of sines states:

The transitive property (switching the orders of the equations) applies here. Therfore, we can say that

We then plug in our given values to find C


Solving, we get 0.8557316387.
We're not done yet!We are trying to find an angle measure, so we'll do the inverse of the ratio we used (sin).
arcsin0.8557316387 (arcsin is the same as inverse sin)
=
58.8 (approximate)
So the measure of angle C is 58.8. You could check this by reinserting it into the equation

.
:)