2. Find the distance between (-3,2) and (4,8).
2 answers:
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<span> For this case we have the following number:
710200
We are going to decompose this number as the sum of three numbers:
</span>
<span> Let's now write each number:
70000: seven hundred thousand
10000: ten thousand
200: two hundred
Therefore, the complete expression is:
Seven hundred ten thousand two hundred
Answer:
Seven hundred ten thousand two hundred</span>
710200
We are going to decompose this number as the sum of three numbers:
</span>
<span> Let's now write each number:
70000: seven hundred thousand
10000: ten thousand
200: two hundred
Therefore, the complete expression is:
Seven hundred ten thousand two hundred
Answer:
Seven hundred ten thousand two hundred</span>
Let's focus on the given equation. The C represents the cost, while the x must be the number of tacos, which is what we're solving for. Since we're given the boundary that C must be 300, then the solution would be:
300 = x² - 40x + 610
x² - 40x + 310 = 0
Apply the quadratic formula where a = 1, b = -40 and c = 310. The roots of x are:
x = 29.49≈30
x = 10.51≈11
<em>Thus, she needs to sell either 30 or 11 tacos.</em>
300 = x² - 40x + 610
x² - 40x + 310 = 0
Apply the quadratic formula where a = 1, b = -40 and c = 310. The roots of x are:
x = 29.49≈30
x = 10.51≈11
<em>Thus, she needs to sell either 30 or 11 tacos.</em>
The weight in 1980 is kilograms
<em><u>Solution:</u></em>
From 1980 to 1990, Lior’s weight increased by 25%
His weight is "k" kilograms in 1990
<em><u>To find: weight in 1980</u></em>
This is a percentage increase problem
Let "x" be the weight in kilograms in 1980
<em><u>The percentage increase is given by formula:</u></em>
Here,
Initial value in 1980 = x
Final value in 1990 = k
Percentage increase = 25 %
<em><u>Substituting the values in formula,</u></em>
Thus the weight in kilograms in 1980 is
Answer:
The coordinates of the explosion is (3300, -2750)
Step-by-step explanation:
I have plot the points and attached to thus answer for easy understanding.
Now, from the question, since station A is the first to hear the explosion, we'll make it the foci of the parabola in the graph I attached and it will be horizontal since the distance between station C and A is much more than that between station B and A. Thus, the reason why station C will have to be the other foci with the hyperbola centred at the origin.
Now, sound travels at a speed of 1100 ft/s and station B is located 1100 ft from station A. Thus, the explosion would likely have occurred at a point on the line x = 3300ft . Since station A is 3300ft from centre C = 3300,hence C² = 3300² = 10,890,000. Since it takes 4 seconds longer for the sound to reach station C than A, the sound has traveled 4(1100)= 4400 ft.
Thus, 4400 = d1 = d2 = 2a
So,2a = 4400 and so, a =2200
a² = 2200² = 4,840,000 where d1 is the distance from station C to the explosion and d2 is the distance from station A to the explosion. To find b², let's use the equation ;
c² = a² + b² and so; b² = c² - a² = 10,890,000 - 4,840,000 = 6,050,000
Equation of hyperbola is given as;
(x²/a²) - (y²/b²) = 1
Plugging in the values of a² and b², we obtain ;
(x²/4,840,000) - (y²/6,050,000) = 1
Since we have deduced that the explosion must occur on the line x= 3300, we'll put in 3300 for x to obtain ;
(3300²/4,840,000) - (y²/6,050,000) = 1
2.25 - 1 = (y²/6,050,000)
y² = (1.25 x 6,050,000)
y² = 7562500
y = √7562500
y = ± 2750
Due to the fact that the explosion will occur at a point further from station B than from station A, the explosion will take place in quadrant 4. Thus, we will take the negative value of y which is - 2750.
So explosion will occur at the coordinate (3300, -2750)
