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3 years ago

What can you conclude about the intersection of a chord and the radius that bisects it

Mathematics

1 answer:

IRINA_888 [86]3 years ago

4 0

Answer: The radius and the cord must be perpendicular to each other.

Step-by-step explanation:

A cord is a line from a point of circumference of a circle to another point of the circumference of the circle without passing through the centre.

A radius is a line from the centre of a circle to the circumference of the circle.

When the radius line intersect and bisects a cord of a circle, the radius and the cord must be perpendicular to each other.

You might be interested in

Find the greatest possible error for each measurement. 10 1/8 oz

scoray [572]

Answer:

0.0005 oz

Step-by-step explanation:

Usually, the greatest number that is allowed for approximation, assuming that the number itself is obtained by approximation, is the greatest possible error of it.

It is normally half the place value of the last digit in a number.

Like here we have $10\frac{1}{8}$ oz which is equal to $10.125$ oz. The last digit is 5 which is at the thousandth place (0.001) so the greatest possible error for this would be its half.

$\frac{0.001}{2}$ = 0.0005 oz

3 0

3 years ago

In a long division exercise the divisor is 8x5−4x. What is the degree of the remainder for which the division process can be st

labwork [276]

Answer:

The degree of the remainder should be 4 for the division process to be stopped

Step-by-step explanation:

From the question, we have the degree of the divisor as 5

So, for the division process to be stopped, the degree of the remainder should be one less than the degree of the divisor

Once the degree of the remainder is less than the degree of the divisor, we have no option that to stop and not proceed further with the division

So in the case of the particular question, the degree of the remainder should be of degree 4

4 0

3 years ago

Solve each system using elimination PLEASE HELP!!!

Mrrafil [7]

1.
$8x+5y=6 \ \ \ |\times 2 \\
3x-2y=10 \ \ \ |\times 5 \\ \\
16x+10y=12 \\
\underline{15x-10y=50} \\
16x+15x=12+50 \\
31x=62 \ \ \ |\div 31 \\
x=2 \\ \\
3x-2y=10 \\
3 \times 2-2y=10 \\
6-2y=10 \ \ \ |-6 \\
-2y=4 \ \ \ |\div (-2) \\
y=-2 \\ \\
(x,y)=(2,-2)$

2.
$-3x+8y=10 \ \ \ |\times 5 \\
5x+6y=80 \ \ \ |\times 3 \\ \\
-15x+40y=50 \\
\underline{15x+18y=240 \ \ } \\
40y+18y=50+240 \\
58y=290 \ \ \ |\div 58 \\
y=5 \\ \\
5x+6y=80 \\
5x+6 \times 5=80 \\
5x+30=80 \ \ \ |-30 \\
5x=50 \ \ \ |\div 5 \\
x=10 \\ \\
(x,y)=(10,5)$

3.
$4x+12y=-24 \\
3x-6y=-18 \ \ \ |\times 2 \\ \\
4x+12y=-24 \\
\underline{6x-12y=-36} \\
4x+6x=-24-36 \\
10x=-60 \ \ \ |\div 10 \\
x=-6 \\ \\
3x-6y=-18 \\
3 \times (-6)-6y=-18 \\
-18-6y=-18 \ \ \ |+18 \\
-6y=0 \ \ \ |\div (-6) \\
y=0 \\ \\
(x,y)=(-6,0)$

4.
$5x-6y=48 \ \ \ |\times 2 \\
2x+5y=-3 \ \ \ |\times (-5) \\ \\
10x-12y=96 \\
\underline{-10x-25y=15} \\
-12y-25y=96+15 \\
-37y=111 \ \ \ |\div (-37) \\
y=-3 \\ \\
2x+5y=-3 \\
2x+5 \times (-3)=-3 \\
2x-15=-3 \ \ \ |+15 \\
2x=12 \ \ \ |\div 2 \\
x=6 \\ \\
(x,y)=(6,-3)$

8 0

3 years ago

Find The range and interquartile 45279.11, 5271. 04, 5309.73, 5363.38, 5409.72, 5444.74, 5603.46, 5695.39, 5871.96, 5972.35

dangina [55]

Answer:

40008 07 is the answer good luck :)

8 0

3 years ago

What are the real roots of 2x^3-42x-40

malfutka [58]

2x^3-42x-40=0
2x^3-42x=40
2x(x^2-21)=40
a. 2x=40=>x=20
b. x^2=21+40=>x^2=61=>x=+-(radical(61)

7 0

3 years ago