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grandymaker [24]
2 years ago
11

"Ready

Mathematics
1 answer:
Ratling [72]2 years ago
3 0

Answer:

sqrt(x^2+y^2)

Step-by-step explanation:

x^2+y^2=z^2 (Pythagorean theorem)

sqrt(x^2+y^2)=sqrt(z^2)

sqrt(x^2+y^2)=z

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Rewrite the equation in standard form y - 3 = 4(x - 3)
kumpel [21]

Answer:

y=4x-9

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Look at the table. Make a conjecture about the sum of the first 20 positive even numbers.
Ludmilka [50]
It will be 20*21 based on the pattern in the chart
3 0
2 years ago
Read 2 more answers
How do you prove each of the following theorems using either a two-column, paragraph, or flow chart proof?
lilavasa [31]

All the theorems are proved as follows.

<h3>What is a Triangle ?</h3>

A triangle is a polygon with three sides , three vertices and three angles.

1. The Triangle sum Theorem

According to the Triangle Sum Theorem, the sum of a triangle's angles equals 180 degrees.

To create a triangle ABC, starting at point A, move 180 degrees away from A to arrive at point B.

We turn 180 degrees from B to C and 180 degrees from C to return to A, giving a total turn of 360 degrees to arrive to A.

180° - ∠A + 180° - ∠B + 180° - ∠C = 360°

- ∠A - ∠B  - ∠C = 360° - (180°+ 180°+ 180°) = -180°

∠A + ∠B  + ∠C = 180°

(Hence Proved)

2. Isosceles Triangle Theorem

Considering an isosceles triangle ΔABC

with AB = AC, we have by sine rule;

\rm \dfrac{sinA}{BC} =  \dfrac{sinB}{AC} =  \dfrac{sinC}{AB}\\

as AB = AC

sin B = sin C

angle B = angle C

3.Converse of the Isosceles theorem

Consider an isosceles triangle ΔABC with ∠B= ∠C, we have by sine rule;

\rm \dfrac{sinA}{BC} =  \dfrac{sinB}{AC} =  \dfrac{sinC}{AB}\\

as  ∠B= ∠C ,

AB = AC

4. Midsegment of a triangle theorem

It states that the midsegment of two sides of a triangle is equal to (1/2)of the third side parallel to it.

Given triangle ABC with midsegment at D and F of AB and AC respectively, DF is parallel to BC

In ΔABC and ΔADF

∠A ≅ ∠A

BA = 2 × DA, BC = 2 × FA

Hence;

ΔABC ~ ΔADF (SAS similarity)

BA/DA = BC/FA = DF/AC = 2

Hence AC = 2×DF

5.Concurrency of Medians Theorem

A median of a triangle is a segment whose end points are on vertex of the triangle and the middle point of the side ,the medians of a triangle are concurrent and  the point of intersection is inside the triangle known as Centroid .

Consider a triangle ABC , X,Y and Z are the midpoints of the sides

Since the medians bisect the segment AB into AZ + ZB

BC into BX + XB

AC into AY + YC

Where:

AZ = ZB

BX = XB

AY = YC

AZ/ZB = BX/XB = AY/YC = 1

AZ/ZB × BX/XB × AY/YC = 1 and

the median segments AX, BY, and CZ are concurrent (meet at point within the triangle).

To know more about Triangle

brainly.com/question/2773823

#SPJ1

8 0
1 year ago
What's the answer to this problem
Grace [21]

Answer:

3' 10'' if I'm not mistaken.

Step-by-step explanation:

5 feet subtracted by 1 foot is 4 feet, 6 inches subtracted by 8 inches sets us back a foot and leaves 10 inches on the previous foot.

7 0
2 years ago
[Q71 Suppose that the height, in inches, of a 25-year-old man is a normal random variable with parameters g = 71 inch and 02 = 6
viktelen [127]

Answer: (a) Percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

              (b) Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

Step-by-step explanation:

Given that,

                  Height (in inches) of a 25 year old man is a normal random variable with mean g=71 and variance o^{2} =6.25.

To find:  (a) What percentage of 25 year old men are 6 feet, 2 inches tall

               (b) What percentage of 25 year old men in the 6 footer club are over 6 feet. 5 inches.

Now,

(a) To calculate the percentage of men, we have to calculate the probability

P[Height of a 25 year old man is over 6 feet 2 inches]= P[X>74in]

                           P[X>74] = P[\frac{X-g}{o} > \frac{74-71}{2.5}]

                                         = P[Z > 1.2]

                                         = 1 - P[Z ≤ 1.2]

                                         = 1 - Ф (1.2)

                                         = 1 - 0.8849

                                         = 0.1151

Thus, percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

(b) P[Height of 25 year old man is above 6 feet 5 inches gives that he is above 6 feet] = P[X, 6ft 5in - X, 6ft]

     P[X > 6ft 5in I X > 6ft] = P[X > 77 I X > 72]

                                          = \frac{P[X > 77]}{P[ X > 72]}

                                          = \frac{P[\frac{X - g}{o}>\frac{77-71}{2.5}]  }{P[\frac{X-g}{o} >\frac{72-71}{2.5}] }

                                          = \frac{P[Z >2.4]}{P[Z>0.4]}

                                          =  \frac{1-P[Z\leq2.4] }{1-P[Z\leq0.4] }

                                          = \frac{1-0.9918}{1-0.6554}

                                          = \frac{0.0082}{0.3446}

                                          = 0.024

Thus, Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

4 0
3 years ago
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