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jekas [21]
4 years ago
5

Who s the first person to eat food?

Mathematics
1 answer:
Blizzard [7]4 years ago
6 0
This is impossible to answer. This all depends on what you define as a 'person.' If you are talking about the first Homo Sapiens, then that would be around 2 million years ago, but the predecessors of Homo sapiens could also be considered 'people,' so then you'd have to determine at what point they came into existence.
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Which expression is equivalent to this polynomial? x2 + 8
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What are the options?
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a DVD box sets includes three trailer movies and two comedies use t2 represent the cost of each trailer and see to represent the
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2t + c = Total cost
HOPE IT HELP!!
4 0
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A cone has a diameter of 18 units and height of 8 units. What is its volume? 72π cubic units 216π cubic units 648π cubic units 8
ANTONII [103]
B 216 because if you multiply pi with the answer you get 678.58
5 0
4 years ago
Read 2 more answers
Help please ITS OF TRIGONOMETRY<br>PROVE ​
Flauer [41]

Answer:

The equation is true.

Step-by-step explanation:

In order to solve this problem, one must envision a right triangle. A diagram used to represent the imagined right triangle is included at the bottom of this explanation.  Please note that each side is named with respect to the angle is it across from.

Right angle trigonometry is composed of a sequence of ratios that relate the sides and angles of a right triangle. These ratios are as follows,

sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}

One is given the following equation,

\frac{sin(A)+sin(B)}{cos(A) +cos(B)}+\frac{cos(A)-cos(B)}{sin(A)-sin(B)}=0

As per the attached reference image, one can state the following, using the right angle trigonometric ratios,

sin(A)=\frac{a}{c}\\\\sin(B)=\frac{b}{c}\\\\cos(A)=\frac{b}{c}\\\\cos(B)=\frac{a}{c}

Substitute these values into the given equation. Then simplify the equation to prove the idenity,

\frac{sin(A)+sin(B)}{cos(A) +cos(B)}+\frac{cos(A)-cos(B)}{sin(A)-sin(B)}=0

\frac{\frac{a}{c}+\frac{b}{c}}{\frac{b}{c}+\frac{a}{c}}+\frac{\frac{b}{c}-\frac{a}{c}}{\frac{a}{c}-\frac{b}{c}}=0

\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{b-a}{c}}{\frac{a-b}{c}}

Remember, any number over itself equals one, this holds true even for fractions with fractions in the numerator (value on top of the fraction bar) and denominator (value under the fraction bar).

\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{b-a}{c}}{\frac{a-b}{c}}

\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{-(a-b)}{c}}{\frac{a-b}{c}}

1+(-1)=0

1-1=0

0=0

7 0
3 years ago
Given y=(2x+3)2, choose the standard form of the given quadratic equation
katrin2010 [14]

y=(2x+3)^2 = 4x^2 +12x +9

Taking 4 out from first two terms, we will get

y=4(x^2+3x)+9

Now we divide the coefficient of x by 2 and add and subtract the square of the result, we will get

y=4(x^2+3x+\frac{9}{4}-\frac{9}{4})+9

Distributing 4

y=4(x+\frac{3}{2})^2+9-9

Cancelling 9

y=4(x+\frac{3}{2})^2

And that's the required standard form.

5 0
3 years ago
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