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elena-14-01-66 [18.8K]
3 years ago
10

There are 6 candidates for student council president, 3 candidates

Mathematics
1 answer:
strojnjashka [21]3 years ago
3 0

Answer:

Step-by-step explanation:multiply then see how may and then boom

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1.What is the equation of the line perpendicular to  that passes through ? Write your answer in slope-intercept form. Show your
Juliette [100K]

Answer:

1. Use a compass to make arc marks which intersect above and below then connect.

2. y=\frac{1}{3}x + 2

Step-by-step explanation:

1. To construct a perpendicular line, use a compass to draw arc marks from one end of the segment through point P. Then repeat this again at the other end. This means at point P there will be two intersecting arc marks. Repeat the process down below with the same radius as used above. Then connect the two intersections.

2. The point slope form of a line is (y-y_1)=m(x-x_1) where x_1=-3\\y_1=1. We write  

(y-1)=m(x--3)\\(y-1)=m(x+3)  

Since the line is to be perpendicular to the line shown it will have the negative reciprocal to the slope of the function 3x+y =-8. To find m, rearrange the function to be y=-8-3x. The slope is -3 and the negative reciprocal will be 1/3.

(y-1)=\frac{1}{3}(x+3)  

Simplify for slope intercept form.

(y-1)=\frac{1}{3}(x+3)\\(y-1)=\frac{1}{3}x+1\\y=\frac{1}{3}x + 2


3 0
3 years ago
Factoring out the GCF of the polynomial 2x^6+2x^5 will give
telo118 [61]
2x^5(x+1)

hope this helps
5 0
3 years ago
Which graph represents x ≥ 2?
emmasim [6.3K]

Answer:

...

Step-by-step explanation:

3 0
3 years ago
Lee ran a mile in 7 1/3 minutes. His friend Sam ran the same mile in 8 5/9 minutes. How many minutes faster did Lee run? (First
Anna [14]

Answer:

1 2/9 minutes faster

Step-by-step explanation:

Take the larger number and subtract the smaller number

8 5/9 minutes - 7 1/3 minutes

Get a common denominator

8 5/9 - 7 1/3 *3/3

8 5/9 - 7 3/9

1 2/9 minutes faster

5 0
3 years ago
Can somebody find the solution to this question
AleksAgata [21]
<h3>Answer: Solution is x = -2</h3>

You have two equations with y1 = f(x) and y2 = g(x).

We're looking for the values of x such that f(x) = g(x). This is the same as trying to solve y1 = y2.

The first row of the table shows y1 and y2 having the same value 5. So we just record the x value that goes with these y values.

5 0
3 years ago
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