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3 years ago

What divides an angle into two congruent angles?

Mathematics

2 answers:

Tpy6a [65]3 years ago

7 0

The answer is Angle Bisectors

a_sh-v [17]3 years ago

3 0

An angle bisector- – formed by a ray going from the midsection of the angle

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F(x)=2x-1 and g(x)=x^2

Fofino [41]

2x-1=x^2

If you plug these two equations into a graphing calculator, like desmos.com, you can see that the two lines will intersect at (1, 1).

5 0

4 years ago

Plz help me with this

Nataly_w [17]

Answer:

x=100

Step-by-step explanation:

100 and x are vertical angles and vertical angles are equal

x = 100

6 0

4 years ago

Read 2 more answers

If f(x) = 3x4 + 2x3 - x + 15, what would be the list of possible rational roots?

Lesechka [4]

Answer: A. ± $\frac{1}{3} ,\frac{5}{3} ,1,3,5,15$

Step-by-step explanation:

± $\frac{1, 3, 5, 15}{1, 3}$

= ± $1, 3, 5, 15, 1/3, 5/3, 5$

Rearranging, we get:

= ± $\frac{1}{3} ,\frac{5}{3} ,1,3,5,15$

7 0

2 years ago

Need help with a math question PLEASE HELP

PolarNik [594]

Answer:

57%

Step-by-step explanation:

From the table you can state that:

the total number of students is 10,730
the number of students receiving Finincial Aid is 6,101

Use the definition of the probability

$Pr=\dfrac{\text{The number of favorable outcomes}}{\text{The number of all possible outcomes}}$

So,

the number of favorable outcomes = 6,101
the number of all possible outcomes = 10,730

Hence, the probability is

$Pr=\dfrac{6,101}{10,730}\approx 0.5685\approx 57\%$

6 0

3 years ago

A cash register contains 5 $10 bills and 3 $5 bills. you randomly pick 4 bills.

kobusy [5.1K]

a. The probability of picking exactly two $10 bills is $\frac{3}{7}$

b. The probability of picking at most one $5 bill is $\frac{1}{2}$

Explanation

Number of $10 bills $= 5$

and number of $5 bills $= 3$

Total number of bills= 5+3 = 8. You need to randomly pick 4 bills, so the total possible outcome $= ^8C^4$

a.

For getting exactly two $10 bills, you need to pick two $10 bills and two $5 bills. That means we will pick two $10 bill from 5 bills and two $5 bills from 3 bills.

So, the probability $=\frac{(^5C^2)*(^3C^2)}{^8C^4} =\frac{10*3}{70}=\frac{30}{70}=\frac{3}{7}$

b.

For getting at most one $5 bill, you need to pick either zero or one $5 bill.

If you pick zero $5 bill, that means there are four $10 bills

and if you pick one $5 bill, that means there are three $10 bills.

So, the probability $=\frac{(^3C^0*^5C^4)+(^3C^1*^5C^3)}{^8C^4}=\frac{(1*5)+(3*10)}{70}=\frac{5+30}{70}=\frac{35}{70}=\frac{1}{2}$

3 0

4 years ago