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mote1985 [20]
2 years ago
14

If m (AB = 120°, then which of the following have measures of 60°?

Mathematics
2 answers:
victus00 [196]2 years ago
7 0
<span>∠BCA = 120 /2 = 60

answer
</span><span>∠BCA</span>
RoseWind [281]2 years ago
4 0
ACD  I  believe is the correct answer
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3 0
2 years ago
In a purse there are 30 coins, twenty one-rupee and remaining 50-paise coins. Eleven coins are picked simultaneously at random a
Masteriza [31]
The solution to the answer is as follows:


<span>0votes</span><span>answered <span>Aug 8, 2015 </span><span><span>by </span>Shashi Kumar Evangelist <span>(3,082<span> points)</span></span></span></span><span>total coins 30
1 rupee coins 20
50 paise coins 10

1 rupee coin probability 20C11
11 coin are picked 30C11
<span>ans should be = 20C11/30C11 = 2/3</span></span>
I hope my answer has come to your help. God bless and have a nice day ahead!
7 0
3 years ago
About 18% of the population of a large country is nervous around strangers
sashaice [31]

Answer:

ok

Step-by-step explanation:

4 0
2 years ago
Twelve times the sum of three and a number is equal to five times the sum of eight more then twice the number
Andreas93 [3]
The equation is 12(3+x)=5(2x+8)
4 0
3 years ago
Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$
Anna [14]

Answer:

\large\boxed{1\dfrac{1}{3}\ u^2}

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}

6 0
3 years ago
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