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3 years ago

TRUE OR FALSE: Every diameter of a circle is a chord. Justify your answer using what you know about diameters and chords.

1 answer:

3 0

True.
A diameter is a line that goes across the whole circle and passes by the center.
A chord is any line segment that connects two points on the circumference.
Since the diameter is a line from one opposite end of the circle to the other, it is connect two points on the circumference, thus it is a chord.

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Two systems of equations are shown below:

Sati [7]

System A:
6x + y = 2
-x - y = -3

System B:
2x - 3y = -10
-x-y = -3

Solve:
System A:
6x + y = 2
y = 2 - 6x
-x - (2-6x) = -3
-x - 2 + 6x = -3
5x = -3 + 2
5x = -1
x = -1/5
y = 2 - 6(-1/5)
y = 2 + 6/5
y = 2 + 1.2
y = 3.2 System A: x = -1/5 or -0.2 ; y = 3 1/5 or 3.2

System B:
2x - 3y = -10
2x = -10 + 3y
x = -5 + 1.5y
-x - y = -3
-(-5 + 1.5y) -y = -3
5 - 1.5y - y = -3
-2.5y = -3 - 5
-2.5y = -8
y = 3.2
x = -5 + 1.5(3.2)
x = -5 + 4.8
x = -0.2 System B: x = -0.2 ; y = 3.2

<span>B) They will have the same solution because the first equations of both the systems have the same graph.</span>

6 0

3 years ago

Read 2 more answers

A group of freshmen at a local university consider joining the equestrian team. Thirty‑five percent of students choose Western r

lesantik [10]

Answer:

a) Option D) 0.75

b) Option D) 0.3

Step-by-step explanation:

We are given the following in the question:

Percentage of students who choose Western riding = 35%

$P(w) = 0.35$

Percentage of students who choose dressage= 45%

$P(d) = 0.45$

Percentage of students who choose jumping = 50%

$P(j) = 0.50$

Percentage of students who choose both dressage and jumping = 20%

$P(d \cap j) = 0.20$

Percentage of students who choose Western and dressage = 10%

$P(w \cap d) = 0.10$

Percentage of students who choose Western and jumping = 0%

$P(w \cap j) = 0$

Thus, we can say

$P(w \cap j \cap d) = 0$

Formula:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

a) P(student chooses dressage or jumping)

$P(d \cup j) = P(d) + P(j) -p(d \cap j)\\= 0.45 + 0.50-0.20 = 0.75$

b) P(student chooses neither dressage nor Western riding)

$= 1 - P(d \cup w)\\= 1 - (P(d) + P(w) - P(d \cap w))\\= 1 - (0.45 + 0.35 - 0.10)= 0.3$

5 0

3 years ago

Mr . Jackson sold 3 computer monitors in the 1st year. In the 2nd year, he sold 4 times as many monitors as the 1st year. In the

tester [92]

So,

1st year: 3
2nd year: 4(first year), which is 12
3rd year: 6(first year), which is 18

The sum of the 1st, 2nd, and 3rd years is 3 + 12 + 18 = 33

Now, multiply 33 by the price at which Mr. Jackson sold each computer ($185).

We get $6105. Mr. Jackson's gross profit was $6105.

3 0

3 years ago

Sorry I forgot to set the last questions as fifty points. Same quesion as before radius, height, and lateral area

Rainbow [258]

R = √10
h = √40 = 2√10
lateral area = 2 pi r h = 2 pi √10 2√10 = 40 pi = 125.6 units^2

4 0

3 years ago

PQ⊥PS , m∠QPR=7x−9, m∠RPS=4x+22<br> Find : m∠QPR

emmainna [20.7K]

Given:

It is given that,

PQ ⊥ PS and