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insens350 [35]
3 years ago
6

Plz find the slope for both of them

Mathematics
1 answer:
masya89 [10]3 years ago
5 0

Answer:

(2,2) : (6,8) → m = 3/2

(-2,5) : (3,-2) → m = -7/5

Explanation:

m (slope) = (y2-y1)/(x2-x1)

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Help!!!
Mrrafil [7]

Answer:

∠ ADC = 53°

Step-by-step explanation:

Note the angle between the radius and the tangent is 90°

The sum of the angles in quadrilateral AOCB = 360°, thus

∠ AOC = 360° - (90 + 90 + 74 )° = 360° - 254° = 106°

∠ AOC is twice the angle on the circumference ∠ ADC, thus

∠ ADC = 106° ÷ 2 = 53°

4 0
3 years ago
Help,anyone can help me do quetion.​
mel-nik [20]
X=60 and Y=30
A quadrilateral with 2 parallel lines is a parallelogram which has equal angle to the opposite angle
3 0
3 years ago
Read 2 more answers
D Cuales son cángulos agudo<br>?​
nasty-shy [4]

Ángulo agudo es aquel que mide menos de 90º

6 0
3 years ago
When comparing prices for activities at a local recreation center , you notice that the center charges different prices on diffe
velikii [3]

We know that charges vary with the days of the week. E.g. Go-karting costs higher ($10) on Saturday and Sunday compared to on Monday to Friday ($5)

We need to know if the cost is a function of the activity type ?

We know that the cost varies with the day. So the cost is a function of the day of the week. On weekdays the function yields a lower value, whereas on weekends it yields a higher value.

Since the cost for a particular activity is changing and not staying constant, <u>hence the cost is not a function of activity type.</u>

8 0
3 years ago
Read 2 more answers
In a class of students, the following data table summarizes how many students have a cat or a dog. What is the probability that
leonid [27]

Given:

Number of students who has a cat and a dog = 5

Number of students who has a cat but do not have a dog = 11

Number of students who has a dog but do not have a cat = 3

Number of students who neither have a cat nor a dog = 2

To find:

The probability that a student has a cat given that they do not have a dog.

Solution:

Let the following events:

A = Student has a cat

B = Do not have a dog

Total number of outcomes is:

5+3+11+2=21

The probability that a student has a cat but do not have a dog is:

P(A\cap B)=\dfrac{11}{21}

The probability that a student do not have a dog is:

P(B)=\dfrac{11+2}{21}

P(B)=\dfrac{13}{21}

The conditional probability is:

P\left(\dfrac{A}{B}\right)=\dfrac{P(A\cap B)}{P(B)}

P\left(\dfrac{A}{B}\right)=\dfrac{\dfrac{11}{21}}{\dfrac{13}{21}}

P\left(\dfrac{A}{B}\right)=\dfrac{11}{13}

Therefore, the probability that a student has a cat given that they do not have a dog is \dfrac{11}{13}.

5 0
3 years ago
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