MATHEMATICS HIGH SCHOOL

A triangle has side lengths of 5, 9 and 12. Is it right acute or obtuse?

Answers

Answer 1
Answer: a\leq b\leq c\ (a;b;c-side\ lengths\ of\ the\ triangle)\n\nthere\ is\ a\ relationship:\n\na^2+b^2=c^2-right\ triangle\n\na^2+b^2 < c^2-obtuse\ triangle\n\na^2+b^2 > c^2-acute\ triangle

a=5;\ b=9;\ c=12\n\na^2+b^2=5^2+9^2=25+81=106\n\nc^2=12^2=144\n\na^2+b^2 < c^2\n\nAnswer:Obtuse\ triangle

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Fill in the missing number in this sequence nine sixteen twenty five forty nine

Answers

9,162,549 Hope this helps not sure if its correct but i think it is :) 

The data set represents the number of rings each person in a room is wearing.0, 2, 4, 0, 2, 3, 2, 8, 6

What is the interquartile range of the data?

2
3
4
6

Answers

Answer:

the interquartile range of the data is:

                                   4

Step-by-step explanation:

We are given a data set as:

             0, 2, 4, 0, 2, 3, 2, 8, 6

On arranging the data in the ascending i.e. increasing order is given by:

    0     0     2     2    2    3      4      6     8

The minimum value of data set=0

Maximum value of data set is: 8

Range of data set= Maximum value-Minimum value

i.e. Range= 8-0

i.e. Range= 8

Also, Median of set is the central tendency of the data and is given by:

Median=  2

Lower set of data is:

                0    0     2    2

Hence, The median of lower set of data is the lower quartile or first quartile.

i.e. Q_1

Hence, Q_1=(0+2)/(2)\n\n\nQ_1=(2)/(2)\n\n\nQ_1=1

Hence, Lower quartile=1

Similarly upper set of data is:

                    3      4      6     8

Hence, The median of upper set of data is the upper quartile or third quartile.

i.e. Q_3

Hence, Q_3=(4+6)/(2)\n\n\nQ_3=(10)/(2)\n\n\nQ_3=5

Hence, Upper quartile=5

Hence, the interquartile range(IQR) is given by:

IQR=Upper quartile-Lower quartile

IQR=5-1

                                    IQR=4
If the data set represents the number of rings each person is wearing, being: 0,2,4,0,2,3,2,8,6, the interquartile range of the data is 2. Being, 4 as the Q1, 3 as the Q2 or median, and 6 as the Q3. Where the formula of getting the interquartile range is IQR= Q1-Q2.

Solve the triangle, find m∠A and m∠C. Round angles to the nearest degree.m∠A= __∘

m∠C= __∘

Answers

In the given right triangle ABC, m∠A ≈ 26.44° and m∠C ≈ 63.56°.

To solve the right triangle ABC, we can use trigonometric ratios. In a right triangle, the three main trigonometric ratios are:

1. Sine (sin): \(\sin(\theta) = \frac{{\text{opposite side}}}{{\text{hypotenuse}}}\)

2. Cosine (cos): \(\cos(\theta) = \frac{{\text{adjacent side}}}{{\text{hypotenuse}}}\)

3. Tangent (tan): \(\tan(\theta) = \frac{{\text{opposite side}}}{{\text{adjacent side}}}\)

Given:

AC = 38

AB = 17

To find the angles m∠A and m∠C, we can use the sine and cosine ratios, respectively.

1. For m∠A:

\(\sin(m\angle A) = \frac{{AB}}{{AC}} = \frac{{17}}{{38}}\)\n\n\(m\angle A= \sin^(-1)\left(\frac{{17}}{{38}}\right)\)

2. For m∠C:

\(\cos(m\angle C) = \frac{{AB}}{{AC}} = \frac{{17}}{{38}}\)\n\n\(m\angle C = \cos^(-1)\left(\frac{{17}}{{38}}\right)\)

Let's calculate the angles:

1. \(m\angle A \approx \sin^(-1)\left(\frac{{17}}{{38}}\right) \approx 26.44^\circ\)\n\n2. \(m\angle C \approx \cos^(-1)\left(\frac{{17}}{{38}}\right) \approx 63.56^\circ\)

Therefore, m∠A ≈ 26.44° and m∠C ≈ 63.56° (rounded to the nearest degree).

To know more about right triangle, refer here:

brainly.com/question/31613708

#SPJ2

Answer:

m\angle A=63^\circ\nm\angle C=26^\circ

Step-by-step explanation:

Trigonometric Ratios

The ratios of the sides of a right triangle are called trigonometric ratios. The longest side of the triangle is called the hypotenuse and the other two sides are called the legs.

Selecting any of the acute angles as a reference, it has an adjacent side and an opposite side. The trigonometric ratios are defined upon those sides.

The cosine ratio is defined as:

\displaystyle \cos\theta=\frac{\text{adjacent leg}}{\text{hypotenuse}}

Note the angle A of the figure has 17 as the adjacent leg and 38 as the hypotenuse, so we can directly apply the formula:

\displaystyle \cos A=(17)/(38)

\cos A=0.4474

Using a scientific calculator, we get the inverse cosine:

A=\arccos(0.4474)

A\approx 63^\circ

Since A+B+C=180°, we can solve for C:

C = 180° - A - B

C = 180° - 63° - 90°

C = 26°

Thus:

m\angle A=63^\circ\nm\angle C=26^\circ

30 POINTS!!!The Pythagorean theorem is used to prove that a triangle is an _____ triangle.

A. Right
B. Isosceles
C. Equilateral
D. Acute

Answers

Answer:

Step-by-step explanation:

A. Right Triangle

If Adam Ct. is perpendicular to Charles st. and Charles St. is parallel to Edward Rd. what must be true?A. Adam ct. is perpendicular to Edward Rd.
B. Adam ct. is parallel to Edward La.
C. Bertha Dr. is parallel to Charles st.
D. Dana la. is perpendicular to Charles st.

Answers

Answer:

A. Adam ct. is perpendicular to Edward Rd.

Step-by-step explanation:

We are given that,

Adam Ct. is perpendicular to Charles St.

Charles St. is parallel to Edward Rd.

So, we get the situation shown below.

It is required to find the relation between Adam Ct. and Edward Rd.

As, we can see that,

Charles St. being parallel to Edward Rd. and Adam Ct. being perpendicular to Charles St.

We get,

Adam Ct. is perpendicular to Edward Rd.

Hence, option A is correct.


Adam Ct and and Edward Rd must be parallel! (answer B). Take a look at the attachment: i drew the plan according to what we know).

(options C and D can also be true, but we don't know enough about the city to verify this)

David's company reimburses his expenses on food, lodging, and conveyance during business trips. The company pays $60 a day for food and lodging and $0.65 for each mile traveled. David drove 600 miles and was reimbursed $3,390.Part A: Create an equation that will determine the number of days x on the trip (3 points)Part B: Solve this equation justifying each step with an algebraic property of equality. (6 points)Part C: How many days did David spend on this trip? (1 points)

Answers

For Part B, Part A of the equation is needed.
Each day for x ($60 food and lodgings).
Miles for y (600 miles)
3390=60x+0.65y
3390=60x+.65(600)
Equation: 3390=60x+390
Subtract 390 from both sides.
3000=60x
Divide 60 on both sides.
50=x

3390=60x+390
3390=60(50)+390
3390=3000+390
3390=3390
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