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777dan777 [17]
3 years ago
12

Simplify (7ty)(3t) (answer fast please)

Mathematics
1 answer:
attashe74 [19]3 years ago
4 0
21t^2y is the answer!
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For the function given, state the starting point for a sample period: f(t)=3cos( πt-1)
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I think its about 30 feet
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4 years ago
Use the equation √4x+15 - 3√x How many potential solutions are there? ​
topjm [15]
There is one potential solutions and no extraneous solutions.
7 0
3 years ago
What is the cost of six filters if 8 filters cost $39.92
KatRina [158]
Alright first off   8 filters cost $39.92 and out goal is to find the cost of   6 filters Set this up as a proportion. 8  = $39.926  = x 
So let find x

x = (6)($39.92) / 8 = $29.94 <span>It costs $29.94 to buy 6 filters.</span>
5 0
4 years ago
Read 2 more answers
A parabola can be drawn given a focus of (5, -3) and a directrix of y=1. Write the equation of the parabola in any form.
Anna71 [15]

Answer:

(x²-10x+33)/(-8) = y

Step-by-step explanation:

The distance between any point on a parabola from both its focus and directrix are the same.

Let's say we have a point (x,y) on the parabola. We can then say that using the distance formula,

\sqrt{(x-5)^2+(y-(-3))^2}is the distance between (x,y) and the focus. Similarly, the distance between (x,y) and the directrix is |y-1| (I use absolute value here because distance is always positive). We can find this equation by taking the shortest distance from the point to the line. Because the closest point to the line will be the same x value as the point itself, the distance is simply the distance between the y value of the point and the y value of the directrix.

Equating the two equations given, we have

\sqrt{(x-5)^2+(y-(-3))^2} = |y-1|

square both sides to get

(x-5)²+(y+3)²=(y-1)²

expand the y components

(x-5)² + y²+6y+9 = y²-2y+1

subtract y²+6y+9 from both sides

(x-5)² = -8y - 8

expand the x components

x²-10x+25 = -8y - 8

add 8 to both sides to isolate the -8y

x²-10x+33 = -8y

divide both sides by -8 to isolate y

(x²-10x+33)/(-8) = y

6 0
3 years ago
There are eight students in a class. Only one of them has passed Exam P/1 and only one of them has passed Exam FM/2. No student
Svetach [21]

Answer: There is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

Step-by-step explanation:

Total number of students = 8

Number of student who has passed Exam P/1 = 1

Number of student who has passed Exam FM/2 = 1

No student has passed more than one exam.

According to question, exactly three students from a randomly chose group of four students have not passed Exam P/1 or Exam FM/2.

So, Probability will be

\frac{^6C_3\times ^2C_1}{^8C_4}\\\\=\frac{20\times 2}{70}\\\\=\frac{4}{7}\\\\=0.57

Hence, there is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

4 0
4 years ago
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